Subject: Optional Mathematics
Let \(\overrightarrow a\) and \(\overrightarrow b\) be two and θ the angle between them, then the scalar product between \(\overrightarrow a\) and \(\overrightarrow b\) is denoted by \(\overrightarrow a\). \(\overrightarrow b\) and is defined as the product of the magnitudes of \(\overrightarrow a\) and \(\overrightarrow b\) with the cosine of the angle θ.
Owing to the different way in which vectors occur in various physical problems, the product of two vectors \(\overrightarrow {a}\) and \(\overrightarrow {b}\) is defined in the following two ways:
The dot product \(\overrightarrow a\) . \(\overrightarrow b\) gives a scalar result while the cross product \(\overrightarrow a\)× \(\overrightarrow b\) gives a vector result.
The scalar product of two vectors \(\overrightarrow a\) and \(\overrightarrow b\) is defined as the product of the magnitude of two vectors multiplied by the cosine of the angle \(\theta\) between their directions.
Thus, \(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos\(\theta\)
where, |\(\overrightarrow a\)| = a and |\(\overrightarrow b\)| = b.
Now,
Draw perpendicular BM from B to OA.
Here,
\(\overrightarrow {OA}\) = \(\overrightarrow a\) and \(\overrightarrow {OB}\) = \(\overrightarrow b\)
Now,
\(\begin{align*} \overrightarrow a . \overrightarrow b &= |\overrightarrow a| |\overrightarrow b| cos \theta\\ &= a b cos\theta\\ &= (OA) (OB) cos\theta\\ &= OA (OB cos\theta)\\ &= (OA)(OM)\\ &= (magnitude\;of\;\overrightarrow a) (component\;of\;\overrightarrow b\;in\;the\;direction\;of\;\overrightarrow a)\\ \end{align*}\)
So, it is clear that the scalar product of two vectors is equivalent to the product of the magnitude of one vector with the component of the other vector in the direction of this vector.
If we write \(\overrightarrow a\) . \(\overrightarrow b\), the rotation of \(\overrightarrow a\) towards \(\overrightarrow b\) is anticlockwise and the angle \(\theta\) is taken to be positive.
∴ \(\overrightarrow a\) . \(\overrightarrow b\) = ab cos\(\theta\)
If we write \(\overrightarrow b\) . \(\overrightarrow a\), the rotation of \(\overrightarrow b\) towards \(\overrightarrow a\) is clockwise and the angle \(\theta\) is taken to be negative.
∴ \(\overrightarrow b\) . \(\overrightarrow a\) = b a cos(-\(\theta\)) = ba cos\(\theta\)
Hence, \(\overrightarrow a\) . \(\overrightarrow b\) = \(\overrightarrow b\) . \(\overrightarrow a\)
Thus, scalar product is commulative.
Let us consider two points A(a1, a2) and B(b1, b2) in the plane. Then,
position vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\)
position vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\) = \(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\)
Magnitudes of \(\overrightarrow a\) and \(\overrightarrow b\) are
|\(\overrightarrow {OA}\)| = OA = a = |\(\overrightarrow a\)|
|\(\overrightarrow {OB}\)| = OB = b = |\(\overrightarrow b\)|
Let \(\angle\)XOA =β, \(\angle\)XOB =α and \(\angle\)AOB =θ. Then,α -β =θ.
Draw perpendiculars AM and BN from A and B to the x-axis. Then,
OM = a1, MA = a2, ON = b1 and NB = b2.
From the right-angled triangle OMA,
cosβ = \(\frac {OM}{OA}\) = \(\frac {a_1}a\)∴ a1 = a cosβ
sinβ = \(\frac {MA}{OA}\) = \(\frac {a_2}a\)∴ a2 = a sinβ
Similarly,
From the right-angled triangle ONB,
b1 = b cosα and b2 = b sinα
Now,
\(\begin{align*} a_1b_1 + a_2b_2 &= a cosβ b cosα + a sinβ sinα\\ &= ab cos (α - β)\\ &= |\overrightarrow a| |\overrightarrow b| cos \theta...................(i)\\ \end{align*}\)
But,
By the deefinition of scalar product of two vectors \(\overrightarrow a\) and \(\overrightarrow b\),
Now,
From (i) and (ii),
\(\overrightarrow a\) . \(\overrightarrow b\) = a1b1 + a2b2.
This result leads us to define the scalar product of two vectors in another way.
Let \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) and \(\overrightarrow b\) = \(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\) be two vectors. Then the scalar product of \(\overrightarrow a\) and \(\overrightarrow b\) is denoted by \(\overrightarrow a\) . \(\overrightarrow b\) and is defined by \(\overrightarrow a\) . \(\overrightarrow b\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) and \(\overrightarrow b\) .\(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\) = a1b1 + a2b2.
Again,
From (i),
|\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = a1b1 + a2b2
or, cos\(\theta\) = \(\frac {a_1b_1 + a_2b_2}{|\overrightarrow a| |\overrightarrow b|}\)
This result gives us angle between two vectors \(\overrightarrow a\) and \(\overrightarrow b\).
\(\theta\) = cos-1 \(\frac {a_1b_1 + a_2b_2}{|\overrightarrow a| |\overrightarrow b|}\)
The following properties are satisfied by the scalar product of vectors:
Let \(\overrightarrow a\), \(\overrightarrow b\) and \(\overrightarrow c\) be three vectors.
Let \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) and \(\overrightarrow b\) = \(\begin{pmatrix} b_1\\ b_2\\ \end{pmatrix}\) be two vectors. If \(\overrightarrow a\) and \(\overrightarrow b\) are perpendicular to each other, then the angle between \(\overrightarrow a\) and \(\overrightarrow b\) is \(\theta\) =- 90°.
Now,
\(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos 90° = 0
Conversely,
Let \(\overrightarrow a\) . \(\overrightarrow b\) = 0
Then,
|\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = 0
or, ab cos\(\theta\) = 0
or, cos\(\theta\) = 0
∴\(\theta\) = 90°
Thus, if two vectors are perpendicular to each other (or orthogonal), their scalar product is zero.
Let \(\overrightarrow a\) and \(\overrightarrow b\) be two vectors. If \(\overrightarrow a\) and \(\overrightarrow b\) are parallel to each other then the angle between them is 0° or 180°.
Now,
If \(\theta\) = 0°, \(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos 0° = ab
If \(\theta\) = 180°, \(\overrightarrow a\) . \(\overrightarrow b\) = |\(\overrightarrow a\)| |\(\overrightarrow b\)| cos\(\theta\) = ab cos 180° = -ab
Thus, two vector \(\overrightarrow a\) and \(\overrightarrow b\) are parallel to each other if \(\overrightarrow a\) . \(\overrightarrow b\) = ab or, \(\overrightarrow a\) . \(\overrightarrow b\) = -ab.
Let \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) be a plane vector.
Then, \(\overrightarrow a\) . \(\overrightarrow a\) = \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) . \(\begin{pmatrix} a_1\\ a_2\\ \end{pmatrix}\) = a1a1 + a2a2= a12 + a22 = a2
∴ a = \(\sqrt {\overrightarrow a. \overrightarrow a}\)
Hence, the length of a vector \(\overrightarrow a\) is the positive square root of the scalar product \(\overrightarrow a\) . \(\overrightarrow a\).
The scalar product of a vector with itself is often written as the square of the vector.
So, \(\overrightarrow a\) . \(\overrightarrow a\) = a2 i.e. \(\overrightarrow {a^2}\) = a2
Let OX and OY be two mutually perpendicular straight lines. Then the unit vector along OX and OY denoted by \(\overrightarrow i\) and \(\overrightarrow j\) are defined by \(\overrightarrow i\) = \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) and \(\overrightarrow j\) = \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\).
Now,
\(\overrightarrow i\) . \(\overrightarrow i\) = \(\overrightarrow {i^2}\) =\(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) . \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) = 1 + 0 = 1
\(\overrightarrow j\) . \(\overrightarrow j\) = \(\overrightarrow {j^2}\) =\(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) . \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) = 0 + 1 = 1
\(\overrightarrow i\) . \(\overrightarrow j\) = \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) . \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) = 0 + 0 = 0
\(\overrightarrow j\) . \(\overrightarrow i\) = \(\begin{pmatrix} 0\\ 1\\ \end{pmatrix}\) . \(\begin{pmatrix} 1\\ 0\\ \end{pmatrix}\) = 0 + 0 = 0
The value of scalar product of \(\overrightarrow i\) and \(\overrightarrow j\) can be remembered from the table given alongside.
Let \(\overrightarrow a\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\) be a vector. It can be written as
\(\overrightarrow a\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\) = \(\begin{pmatrix} x\\ 0\\ \end{pmatrix}\) = \(\begin{pmatrix} 0\\ y\\ \end{pmatrix}\) = x\(\begin{pmatrix}1\\ 0\\ \end{pmatrix}\) + y\(\begin{pmatrix}0\\ 1\\ \end{pmatrix}\) = x\(\overrightarrow i\) + y\(\overrightarrow j\).
Similarly,
If \(\overrightarrow a\) = nx + y\(\overrightarrow j\), then:
\(\overrightarrow a\)= x\(\overrightarrow i\) + y\(\overrightarrow j\) = x\(\begin{pmatrix}1\\ 0\\ \end{pmatrix}\) + y\(\begin{pmatrix}0\\ 1\\ \end{pmatrix}\) = \(\begin{pmatrix}x\\ 0\\ \end{pmatrix}\) + \(\begin{pmatrix}0\\ y\\ \end{pmatrix}\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\)
Hence, every plane vector \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\) can be represented by x\(\overrightarrow i\) + y\(\overrightarrow j\) and conversely.
Let \(\overrightarrow a\) = a1\(\overrightarrow i\) + a2\(\overrightarrow j\) and \(\overrightarrow b\) = b1\(\overrightarrow i\) + b2\(\overrightarrow j\)
Let \(\overrightarrow a\) = x\(\overrightarrow i\) + y\(\overrightarrow j\) be a vector.
Then,
\(\overrightarrow a\) = \(\begin{pmatrix}x\\ y\\ \end{pmatrix}\)
∴ X- component of \(\overrightarrow a\) = x and Y-component of \(\overrightarrow a\) = y
Magnitude of \(\overrightarrow a\)
|\(\overrightarrow a\)| = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(coefficient\;of\;\overrightarrow i)^2 + (coefficient\;of\;\overrightarrow j)^2}\)
Direction of \(\overrightarrow a\)
tan\(\theta\) = \(\frac yx\) = \(\frac {coefficient\;of\;\overrightarrow j}{coefficient\;of\;\overrightarrow i}\)
Unit vector along the direction of \(\overrightarrow a\)
\(\widehat a\) = \(\frac {\overrightarrow a}{|\overrightarrow a|}\) = \(\frac {1}{\sqrt {x^2 + y^2}}\) (x\(\overrightarrow i\) + y\(\overrightarrow j\)) = \(\frac {x\overrightarrow i + y\overrightarrow j}{\sqrt {x^2 + y^2}}\)
Theorem 1: (Mid- point Formula)
If \(\overrightarrow a\) and \(\overrightarrow b\) are position vector of two points A and B respectively and M is the middle point of the line segment AB, then the position vector of M is \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\)).
Proof:
Let AB be a line segment and O be the origin.
Here,
Position Vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\)
Position Vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\)
Let M be the middle point of the segment AB.
Then,
\(\begin{align*} \overrightarrow {OM} &= \overrightarrow {OA} + \overrightarrow {AM}\\ &= \overrightarrow {OA} + \frac 12\overrightarrow {AB}\\ &= \overrightarrow {OA} + \frac 12(\overrightarrow {OB} - \overrightarrow {OA})\\ &= \overrightarrow a + \frac 12(\overrightarrow b - \overrightarrow a)\\ &= \frac {2\overrightarrow a + \overrightarrow b - \overrightarrow a}{2}\\ &= \frac 12(\overrightarrow a + \overrightarrow b)\\ \end{align*}\)
∴ Position Vector of M = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\)) Proved
Theorem 2: (Section Formula for Internal Division)
If \(\overrightarrow a\) and \(\overrightarrow b\) are position vector of two points A and B respectively and the point M divides the line segment AB internally in the ratio m : n, then the position vector of M is \(\frac {m \overrightarrow b + n\overrightarrow a}{m + n}\).
Proof:
Let AB be a line segment and O be the origin.
Here,
Position Vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\)
Position Vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\)
Let the point M divides AB internally in the ratio m : n.
Then,
\(\begin{align*} \overrightarrow {OM} &= \overrightarrow {OA} + \overrightarrow {AM}\\ &= \overrightarrow {OA} + \frac {m}{m + n}\overrightarrow {AB}\\ &= \overrightarrow {OA} + \frac m{m + n} (\overrightarrow {OB} - \overrightarrow {OA})\\ &= \overrightarrow a + \frac m{m + n} (-\overrightarrow a)\\ &= \frac {m\overrightarrow a + n\overrightarrow a + m\overrightarrow b - m\overrightarrow a}{m + n}\\ &= \frac {m\overrightarrow b + n\overrightarrow a}{m + n}\\ \end{align*}\)
∴ Position Vector of M = \(\frac {m\overrightarrow b + n\overrightarrow a}{m + n}\) Proved
Theorem 3: (Section Formula for External Division)
If \(\overrightarrow a\) and \(\overrightarrow b\) are position vector of two points A and B respectively and the point P divides the line segment AB externally in the ratio m : n, then the position vector of P is \(\frac {m \overrightarrow b - n\overrightarrow a}{m - n}\).
Proof:
Let AB be a line segment and O be the origin.
Here,
Position Vector of A = \(\overrightarrow {OA}\) = \(\overrightarrow a\)
Position Vector of B = \(\overrightarrow {OB}\) = \(\overrightarrow b\)
Let the point P divides AB externally in the ratio m : n.
Then,
\(\frac {AP}{BP}\) = \(\frac mn\)
∴ n\(\overrightarrow {AP}\) = m\(\overrightarrow {BP}\)
or, n (\(\overrightarrow {OP}\) - \(\overrightarrow {OA}\)) = m(\(\overrightarrow {OP}\) - \(\overrightarrow {OB}\))
or, n\(\overrightarrow {OP}\) - n\(\overrightarrow {OA}\) = m\(\overrightarrow {OP}\) - m\(\overrightarrow {OB}\)
or, m\(\overrightarrow {OB}\) - n\(\overrightarrow {OA}\) = m\(\overrightarrow {OP}\) - n\(\overrightarrow {OP}\)
or, m\(\overrightarrow b\) - n\(\overrightarrow a\) = (m - n)\(\overrightarrow {OP}\)
∴ \(\overrightarrow {OP}\) = \(\frac {m\overrightarrow b - n\overrightarrow a}{m - n}\)
So, the position vector of P = \(\frac {m\overrightarrow b - n\overrightarrow a}{m - n}\) Proved
Theorem 4:
The line segment joining the mid-point of two sides of a triangle is parallel to the third side and it is half of it.
Proof:
Let ABC be a triangle and P and Q be the mid-points of the sides AB and AC respectively.
Here,
\(\overrightarrow {PA}\) = \(\frac 12\)\(\overrightarrow {BA}\)
\(\overrightarrow {AQ}\) = \(\frac 12\)\(\overrightarrow {AC}\)
Then,
\(\overrightarrow {PA}\) + \(\overrightarrow {AQ}\) = \(\frac 12\)\(\overrightarrow {BA}\) + \(\frac 12\)\(\overrightarrow {AC}\)
∴ \(\overrightarrow {PQ}\) = \(\frac 12\)(\(\overrightarrow {BA}\) + \(\overrightarrow {AC}\))
i.e. \(\overrightarrow {PQ}\) = \(\frac 12\)\(\overrightarrow {BC}\)
Clearly, \(\overrightarrow {PQ}\) // \(\overrightarrow {BC}\) Proved
Theorem 5:
The position vector of the centroid of a triangle is given by \(\overrightarrow g\) = \(\frac 13\) (\(\overrightarrow a\) + \(\overrightarrow b\) + \(\overrightarrow c\)) where \(\overrightarrow a\), \(\overrightarrow b\) and \(\overrightarrow c\) are the position vectors of the vertices and \(\overrightarrow g\) is the position vector of the centroid.
Proof:
Let ABC be triangle and O be the origin.
Let,
\(\overrightarrow {OA}\) = \(\overrightarrow a\)
\(\overrightarrow {OB}\) = \(\overrightarrow b\)
\(\overrightarrow {OC}\) = \(\overrightarrow c\)
Let D be the mid- point of AC.
By mid-point theorem,
\(\overrightarrow {OD}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow c\))
Let G be the centroid of the triangle ABC.
Then G divides BD internally in the ratio 2: 1.
Then,
\(\begin{align*} \overrightarrow {OG} &= \frac {2\overrightarrow {OD} + 1\overrightarrow {OB}}{2 + 1}\\ &= \frac {2 × \frac 12 \overrightarrow a + \overrightarrow c + \overrightarrow b}{3}\\ &= \frac 13 (\overrightarrow a + \overrightarrow b + \overrightarrow c)\\ \end{align*}\)
∴ Position Vector of G (\(\overrightarrow g\)) = \(\frac 13 (\overrightarrow a + \overrightarrow b + \overrightarrow c)\) Proved
Theorem 6:
The median to the base of an isosceles triangle is perpendicular to the base.
ABC be an isosceles triangle where AB = ACand AM is the median to the base BC.Proof:
Let \(\overrightarrow {AB}\) = \(\overrightarrow a\) and \(\overrightarrow {AC}\) = \(\overrightarrow b\)
Then,
|\(\overrightarrow a\)| = |\(\overrightarrow b\)| or a = b
Now,
\(\overrightarrow {AM}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {BM}\) = \(\overrightarrow {AB}\) + \(\frac 12\)\(\overrightarrow {BC}\)
\(\overrightarrow {BC}\) = \(\overrightarrow {BA}\) + \(\overrightarrow {AC}\) = -\(\overrightarrow a\) + \(\overrightarrow b\)
\(\begin{align*} \therefore \overrightarrow {AM} &= \overrightarrow a + \frac 12 (\overrightarrow b - \overrightarrow a)\\ &= \frac {\overrightarrow a + \overrightarrow b - \overrightarrow c}2\\ &= \frac 12(\overrightarrow a + \overrightarrow b)\\ \end{align*}\)
Now,
\(\begin{align*} \overrightarrow {AM} . \overrightarrow {BC} &= \frac 12(\overrightarrow a + \overrightarrow b) . (\overrightarrow b - \overrightarrow a)\\ &= \frac 12 (\overrightarrow b + \overrightarrow a) . (\overrightarrow b - \overrightarrow a)\\ &= \frac 12(b^2 - a^2)\\ &= 0\\ \end{align*}\)
∴ AM⊥ BC Proved
Theorem 7:
The figure formed by joining the mid-point of the adjacent sides of a quadrilateral is a parallelogram.
Proof:
Let ABCD be a quadrilateral and P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Join BD.
Then,
\(\overrightarrow {BA}\) + \(\overrightarrow {AD}\) = \(\overrightarrow {BD}\)
2\(\overrightarrow {PA}\) + 2\(\overrightarrow {AS}\) = \(\overrightarrow {BD}\)
\(\overrightarrow {PA}\) + \(\overrightarrow {AS}\) = \(\frac 12\)\(\overrightarrow {BD}\)
∴ \(\overrightarrow {PS}\) = \(\frac 12\)\(\overrightarrow {BD}\)
Clearly, \(\overrightarrow {PS}\) // \(\overrightarrow {BD}\)
Again,
\(\overrightarrow {BC}\) + \(\overrightarrow {CD}\) = \(\overrightarrow {BD}\)
2\(\overrightarrow {QC}\) + 2\(\overrightarrow {CR}\) = \(\overrightarrow {BD}\)
\(\overrightarrow {QC}\) + \(\overrightarrow {CR}\) = \(\frac 12\)\(\overrightarrow {BD}\)
∴ \(\overrightarrow {QC}\) = \(\frac 12\)\(\overrightarrow {BD}\)
Clearly, \(\overrightarrow {QR}\) // \(\overrightarrow {BD}\)
\(\overrightarrow {PS}\) = \(\overrightarrow {QR}\), \(\overrightarrow {PS}\) // \(\overrightarrow {QR}\)
Similarly,
\(\overrightarrow {PQ}\) = \(\overrightarrow {SR}\), \(\overrightarrow {PQ}\) // \(\overrightarrow {SR}\)
Hence, PQRS is a parallelogram.
Theorem 8:
The diagonals of a parallelogram bisect each other.
Proof:
Let OACB be a parallelogram and O be the origin.
Let \(\overrightarrow {OA}\) = \(\overrightarrow a\) and \(\overrightarrow {OB}\) = \(\overrightarrow b\)
Let M be the middle point of AB.
Then,
\(\overrightarrow {OM}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\)).............................................(i)
Also,
\(\overrightarrow {OC}\) = \(\overrightarrow a\) + \(\overrightarrow b\), by parallelogram of vector addition.
Let N be the middle point of OC.
Then,
\(\overrightarrow {ON}\) = \(\frac 12\)\(\overrightarrow {OC}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\))............................................(ii)
From (i) and (ii), M and N have the same position vector. So, M and N are same points.
Hence, the diagonals of a parallelogram bisect each other.
Theorem 9:
The diagonals of a rectangle are equal.
Proof:
Let ABCD be a rectangle and AC and BD are diagonals.
Then,
\(\overrightarrow {AB}\) = \(\overrightarrow {DC}\), \(\overrightarrow {AD}\) = \(\overrightarrow {BC}\), \(\angle A\) = \(\angle B\) = \(\angle C\) = \(\angle D\) = 90° (\(\overrightarrow {AB}\) . \(\overrightarrow {BC}\) = 0, \(\overrightarrow {BA}\) . \(\overrightarrow {AD}\) = 0)
Now,
\(\overrightarrow {AC}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {BC}\)
\(\overrightarrow {AC}^2\) = (\(\overrightarrow {AB}\) + \(\overrightarrow {BC}\))2
AC2 = \(\overrightarrow {AB}^2\) + 2\(\overrightarrow {AB}\) . \(\overrightarrow {BC}\) + \(\overrightarrow {BC}^2\)
AC2= AB2 + 0 + BC2
AC2= AB2 + BC2......................................(i)
Again,
\(\overrightarrow {BD}\) = \(\overrightarrow {BA}\) + \(\overrightarrow {AD}\)
\(\overrightarrow {BD}^2\) = (\(\overrightarrow {BA}\) + \(\overrightarrow {AD}\))2
BD2= \(\overrightarrow {BA}^2\) + 2\(\overrightarrow {BA}\) . \(\overrightarrow {AD}\) + \(\overrightarrow {AD}^2\)
BD2= BA2 + 0 + AD2
BD2= AB2 + BC2......................................(i)
From (i) and (ii),
AC2= BD2
∴ AC = BD
Hence, the diagonals of a rectangle are equal.
Theorem 10:
The diagonals of a rhombus bisect each other at right angles.
Proof:
Let ABCD is a rhombus and AC and BD are diagonals.
Obviously,
\(\overrightarrow {AO}\) = \(\overrightarrow {OC}\) and \(\overrightarrow {BO}\) = \(\overrightarrow {OD}\)
Now,
We prove that,
\(\overrightarrow {AO}\)⊥ \(\overrightarrow {BO}\)
Let \(\overrightarrow {AB}\) = \(\overrightarrow a\) and \(\overrightarrow {AD}\) = \(\overrightarrow b\)
Then,
\(\overrightarrow {AC}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {BC}\) = \(\overrightarrow {AB}\) + \(\overrightarrow {AD}\) = \(\overrightarrow a\) + \(\overrightarrow b\)
\(\overrightarrow {AO}\) = \(\frac 12\)\(\overrightarrow {AC}\) = \(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\))..............................(i)
Again,
\(\overrightarrow {BD}\) = \(\overrightarrow {BA}\) + \(\overrightarrow {AD}\) = -\(\overrightarrow a\) + \(\overrightarrow b\) = \(\overrightarrow b\) - \(\overrightarrow a\)
\(\overrightarrow {BO}\) = \(\frac 12\)\(\overrightarrow {BD}\) = \(\frac 12\)(\(\overrightarrow b\) - \(\overrightarrow a\))..............................(ii)
From (ii) and (ii),
\(\overrightarrow {AO}\) .\(\overrightarrow {BO}\) =\(\frac 12\)(\(\overrightarrow a\) + \(\overrightarrow b\) .\(\frac 12\)(\(\overrightarrow b\) - \(\overrightarrow a\)) = \(\frac 14\) (\(\overrightarrow b^2\) - \(\overrightarrow a^2\)) = \(\frac 14\)(b2 - a2) = 0
Hence, the diagonals of a rhombus bisect each other at right angles.
Theorem 11:
The angle in a semi- circle is a right angle.
Proof:
Let O be the centre of a circle and AB be a diameter. \(\angle\)ACB is an angle in the semi- circle. Join OC.
Now,
\(\overrightarrow {AC}\) = \(\overrightarrow {AO}\) + \(\overrightarrow {OC}\)..............................(i)
\(\overrightarrow {CB}\) = \(\overrightarrow {CO}\) + \(\overrightarrow {OB}\) = \(\overrightarrow {CO}\) + \(\overrightarrow {AO}\)
\(\overrightarrow {CB}\) = \(\overrightarrow {AO} - \overrightarrow {OC}\)...................................(ii)
From (i) and (ii),
\(\begin{align*} \overrightarrow {AC} . \overrightarrow {CB} &= (\overrightarrow {AO} + \overrightarrow {OC}) . (\overrightarrow {AO} - \overrightarrow {OC})\\ & =(\overrightarrow {AO}^2 - \overrightarrow {OC}^2\\ &= AO^2 - OC^2 [since\; AO = OC]\\ &= 0\\ \end{align*}\)
Hence, the angle in the semi- circle is a right angle.
Theorem 12:
The mid- point of the hypotenuse of a right-angled triangle is equidistant from its vertices.
Proof:
Let AOC be a right angled triangle and O be the origin.
Let \(\angle AOC\) = 90° and M is the mid- point of AC.
Then,
\(\overrightarrow {OM}\) = \(\frac {\overrightarrow {OA} + \overrightarrow {OC}}2\)
2\(\overrightarrow {OM}\) = \(\overrightarrow {OA}\) + \(\overrightarrow {OC}\)
(2\(\overrightarrow {OM}\))2 = (\(\overrightarrow {OA}\) + \(\overrightarrow {OC}\))2
4 \(\overrightarrow {OM}^2\) = \(\overrightarrow {OA}^2\) + 2\(\overrightarrow {OA}\) . \(\overrightarrow {OC}\) + \(\overrightarrow {OC}^2\)
4 OM2 = OA2 + 0 + OC2
4 OM2 = OA2 + OC2
4 OM2 = AC2
OM2 = \(\frac {AC^2}4\)
∴ OM = \(\frac 12\)AC
Also,
AM = MC = \(\frac 12\)AC
∴ AM = MC = OM
Hence, the mid- point of the hypotenuse of a right angled triangle is equidistant from its vertices.
Dot product of \(\overrightarrow a\) & \(\overrightarrow b\) = \(\overrightarrow a\) . \(\overrightarrow b\)
If \(\overrightarrow i\) and \(\overrightarrow j\) are unit vectors along x-axis and y-axis \(\overrightarrow i\).\(\overrightarrow i\) = \(\overrightarrow j\). \(\overrightarrow j\) = 1 and \(\overrightarrow i\).\(\overrightarrow j\) = \(\overrightarrow j\).\(\overrightarrow i\) = 0
If \(\vec a\) = \(\begin {pmatrix} 6\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} -1\\ 6\\ \end {pmatrix}\), then prove that:
\(\vec a\) and \(\vec b\) are perpendiculsr to each other.
Here,
\(\vec a\) = \(\begin {pmatrix} 6\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} -1\\ 6\\ \end {pmatrix}\)
\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 6× -1 + 1× 6 = 6 - 6 = 0
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(6)^2 + (1)^2}\) = \(\sqrt {36 + 1}\) = \(\sqrt {37}\)
\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(-1)^2 + (6)^2}\) = \(\sqrt {1 + 36}\) = \(\sqrt {37}\)
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)
or, cos\(\theta\) = \(\frac 0{\sqrt {37} \sqrt {37}}\)
or, cos\(\theta\) = 0
or, \(\theta\) = cos-1(0) = 90°
The angle between two vectors is 90° so, the \(\vec a\) and \(\vec b\) are perpendicular each other. Proved
If \(\vec a\) = \(\begin {pmatrix} 2\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} 0\\ -2\\ \end {pmatrix}\), find the angle between \(\vec a\) and \(\vec b\).
Here,
\(\vec a\) = \(\begin {pmatrix} 2\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} 0\\ -2\\ \end {pmatrix}\)
\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 2 × 0 + 1 × -2 = 0 + -2 = -2
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(2)^2 + (1)^2}\) = \(\sqrt {4 + 1}\) = \(\sqrt 5\)
\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(0)^2 + (-2)^2}\) = \(\sqrt {0 + 4}\) = 2
cos\(\theta\) = \(\frac {-2}{2\sqrt 5}\) = -\(\frac 1{\sqrt 5}\)
∴ \(\theta\) = cos-1 (-\(\frac 1{\sqrt 5}\)) Ans
If \(\vec p\) = \(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\) and \(\vec q\) = \(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\), find the angle between \(\vec p\) and \(\vec q\).
Here,
\(\vec p\) = \(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\) and \(\vec q\) = \(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\)
\(\vec p\) . \(\vec q\) = x1x2 + y1y2 = 3 × 1 + 4 × 1 = 3 + 4 = 7
\(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt 25\) = 5
\(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(1)^2 + (1)^2}\) = \(\sqrt {1 + 1}\) = \(\sqrt 2\)
cos\(\theta\) = \(\frac {7}{5\sqrt 2}\) = \(\frac 7{7.07}\) = 0.99
∴ \(\theta\) = cos-1 (0.99) = 8.110Ans
If \(\vec a\) = \(\begin {pmatrix} 2\\ -3\\ \end {pmatrix}\) and \(\vec c\) = \(\begin {pmatrix} 4\\ -2\\ \end {pmatrix}\), find the angle between \(\vec a\) and \(\vec c\).
Here,
\(\vec a\) = \(\begin {pmatrix} 2\\ -3\\ \end {pmatrix}\) and \(\vec c\) = \(\begin {pmatrix} 4\\ -2\\ \end {pmatrix}\)
\(\vec a\) . \(\vec c\) = x1x2 + y1y2 = 2× 4+ (-3) × (-2) = 8+ 6=14
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(2)^2 + (-3)^2}\) = \(\sqrt {4 + 9}\) = \(\sqrt {13}\)
\(\begin {vmatrix} \vec c\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(4)^2 + (-2)^2}\) = \(\sqrt {16 + 4}\) = \(\sqrt {20}\)
cos\(\theta\) = \(\frac {14}{\sqrt {13} . \sqrt {20}}\) = \(\frac {14}{16.13}\) = 0.87
∴ \(\theta\) = cos-1 (0.87) = 29.540Ans
If \(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = 3, \(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = 3\(\sqrt 2\) and \(\vec p\).\(\vec q\) = 9, find the angle between \(\vec p\) and \(\vec q\).
Here,
\(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = 3, \(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = 3\(\sqrt 2\) and \(\vec p\).\(\vec q\) = 9
We know,
cos\(\theta\) = \(\frac {\vec p . \vec q}{\begin {vmatrix} \vec p\\ \end {vmatrix}\begin {vmatrix} \vec q\\ \end {vmatrix}}\) = \(\frac 9{3 × 3\sqrt 2}\) = \(\frac 1{\sqrt 2}\)
cos\(\theta\) = \(\frac 1{\sqrt 2}\)
\(\theta\) = cos-1(\(\frac 1{\sqrt 2}\)) = 45°
∴ The angle between \(\vec p\) and \(\vec q\) is 45°. Ans
If \(\vec a\) = 4 \(\vec i\) + \(\vec j\) and \(\vec b\) = 3 \(\vec i\) + 4\(\vec j\):
Here,
\(\vec a\) = 4 \(\vec i\) + \(\vec j\) and \(\vec b\) = 3 \(\vec i\) + 4\(\vec j\)
(i)
\(\vec a\).\(\vec b\) = (4 \(\vec i\) + \(\vec j\)) . (3\(\vec i\) + 4\(\vec j\)) = 4× 3 + 1× 4 = 12 + 4 = 16 Ans
(ii)
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(4)^2 + (1)^2}\) = \(\sqrt {16 + 1}\) = \(\sqrt {17}\)
\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5
Let: \(\theta\) be the angle between \(\vec a\) and \(\vec b\) then,
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac {16}{5\sqrt {17}}\)
\(\theta\) = cos-1(\(\frac {16}{5\sqrt {17}}\)) = 37.47°
∴ The angle between \(\vec a\) and \(\vec b\) is 37.47°. Ans
Prove that: \(\vec a\) = -4\(\vec i\) + 5\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 10\(\vec j\) are parallel to each other.
Here,
\(\vec a\) = -4\(\vec i\) + 5\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 10\(\vec j\)
(x1 , y1) = (-4 , 5) and (x2 , y2) = (8 , -10)
\(\vec a\).\(\vec b\) = x1x2 + y1y2 = -4 × 8 + 5 × 10 = -32 - 50 = -82
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(-4)^2 + (5)^2}\) = \(\sqrt {16 + 25}\) = \(\sqrt {41}\)
\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-10)^2}\) = \(\sqrt {64 + 100}\) = \(\sqrt {164}\)
We know,
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac {-82}{\sqrt {41} \sqrt {164}}\) = - 1 (Approx.)
\(\theta\) = cos-1 (-1) = 180°
∴ The \(\vec a\) and \(\vec b\) are parallel in opposite direction. Proved
If \(\vec a\) = 3 + k and \(\vec b\) = -7 + 3 are perpendicular each other, find the value of k.
Here,
\(\vec a\) = 3 + k and \(\vec b\) = -7 + 3 are perpendiculat each other.
Then:
\(\vec a\) . \(\vec b\) = 0
or, (3 + k) (-7 + 3) = 0
or, 3× -7 + k× 3 = 0
or, -21 + 3k = 0
or, 3k = 21
or, k = \(\frac {21}3\)
∴ k = 7 Ans
Prove that: \(\vec a\) and \(\vec b\) are perpendicular, if \(\vec a\) = 4 + 2 and \(\vec b\) = -1 + 2.
Here,
\(\vec a\) = 4 + 2 and \(\vec b\) = -1 + 2
If \(\vec a\) and \(\vec b\) are perpendicular than \(\vec a\) . \(\vec b\) = 0
or, (4 + 2) (-1 + 2) = 0
or, 4× -1 + 2× 2 = 0
or, -4 + 4 = 0
∴ 0 = 0
∴ \(\vec a\) and \(\vec b\) are perpendicular to each other. Proved
If \(\vec p\) = 10\(\vec i\) + 2k \(\vec j\) and \(\vec q\) = 2\(\vec i\) - 5\(\vec j\) are perpendicular to each other, find the value of k.
Here,
\(\vec p\) = 10\(\vec i\) + 2k \(\vec j\) and \(\vec q\) = 2\(\vec i\) - 5\(\vec j\)
(x1, y1) = (10, 2k) and (x2, y2) = (2. -5)
We know,
\(\vec p\) and \(\vec q\) are perpendicular.
Then: \(\vec p\) . \(\vec q\) = 0
or, (10, 2k) (2, -5) = 0
or, 10× 2 + 2k× -5 = 0
or, 20 - 10k = 0
or, 10k = 20
or, k = \(\frac {20}{10}\)
∴ k = 2 Ans
If \(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\), find the value of \(\angle\)AOB.
Here,
\(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\)
\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 7 × 5 + (-5) × (-7) = 35 + 35 = 70
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(7)^2 + (-5)^2}\) = \(\sqrt {49 + 25}\) = \(\sqrt {74}\)
\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(5)^2 + (-7)^2}\) = \(\sqrt {25 + 49}\) = \(\sqrt {74}\)
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) =\(\frac {70}{\sqrt {74} \sqrt {74}}\) = \(\frac {70}{74}\) = 0.95
\(\theta\) = cos-1(0.95) = 18.93°
∴\(\angle\)AOB = 18.93° Ans
In what condition \(\vec a\) and \(\vec b\) are parallel? If \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 6\(\vec j\), find the angle between the vectors \(\vec a\) and \(\vec b\).
Let: the vector \(\vec a\) and \(\vec b\) are parallel if \(\vec a\) = m\(\vec b\) where m is scalar quantity.
\(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and b = 8\(\vec i\) - 6\(\vec j\)
\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 3 × 8 + 4 × (-6) = 24 - 24 = 0
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5
\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-6)^2}\) = \(\sqrt {64 + 36}\) = \(\sqrt {100}\) = 10
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac 0{5 × 10}\) = 0
∴ \(\theta\) = cos-1 (0) = 90° Ans
In the given figure, \(\vec {OA}\) = \(\vec a\) and \(\vec {OB}\) = \(\vec b\). If \(\vec {AC} \) = 3 \(\vec {AB}\), find \(\vec {OC}\).
Here,
\(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and \(\vec {AC}\) = 3\(\vec {AB}\) then, \(\vec {OC}\) = ?
Using triangle law in vector addition,
\(\vec {OA}\) + \(\vec {AB}\) = \(\vec {OB}\)
\(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) = \(\vec b\) - \(\vec a\)
We know,
\(\vec {AC}\) = 2\(\vec {AB}\) = 3 (\(\vec b\) - \(\vec a\)) = 3\(\vec b\) - 3\(\vec a\)
Again,
\(\vec {OA}\) + \(\vec {AC}\) = \(\vec {OC}\)
∴ \(\vec {OC}\) = \(\vec a\) + 3\(\vec b\) - 3\(\vec a\) = 3\(\vec b\) - 2\(\vec a\) Ans
If A(1, 2) and B(3, 0) are two points. Express in the form x + y.
Given points are: A(1, 2) and B(3, 0)
x-component = x2 - x1 = 3 - 1 = 2
y-component = y2 - y1 = 0 - 2 = - 2
\(\vec {AB}\) = (2, -2)
\(\vec {AB}\) = 2\(\vec i\) - 2\(\vec j\) Ans
If C(1, 1) and D(-2, -4) are two points. Express in the form of x + y.
Given points are: C(1, 1) and D(-2, -4)
x-component = x2 - x1 = -2 - 1 = -3
y-component = y2 - y1 = 0 - 2 = - 2
\(\vec {CD}\) = (-3, -5)
\(\vec {CD}\) = -3\(\vec i\) - 5\(\vec j\) Ans
Find the position vector of the mid-point M of the line segment PQ. If the co-ordinates of the points P and Q are (3, 5) and (-7, 3) respectively.
Given points are: P(3, 5) and Q(-7, 3).
Let, O be the origin.
\(\vec {OP}\) = \(\begin {pmatrix} 3\\ 5\\ \end {pmatrix}\) and \(\vec {OQ}\) = \(\begin {pmatrix} -7\\ 3\\ \end {pmatrix}\)
Using midpoint formula,
\(\vec {OM}\) = \(\frac {\vec {OP} + \vec {OQ}}2\)
or, \(\vec {OM}\) = \(\frac {\begin {pmatrix}3\\ 5\\ \end{pmatrix} + \begin {pmatrix} -7\\ 3\\ \end {pmatrix}}2\)
or, \(\vec {OM}\) = \(\frac {\begin {pmatrix}3 - 7\\ 5 + 3\\ \end {pmatrix}}2\)
or, \(\vec {OM}\) = \(\frac {\begin {pmatrix} -4\\ 3\\ \end {pmatrix}}2\)
or, \(\vec {OM}\) = \(\begin {pmatrix} \frac {-4}2\\ \frac 82\\ \end {pmatrix}\)
∴ \(\vec {OM}\) = \(\begin {pmatrix} -2\\ 4\\ \end {pmatrix}\)
∴ \(\vec {OM}\) = -2\(\vec i\) + 4\(\vec j\) Ans
In \(\triangle\)ABC , if D is the midpoint of BC, then prove by vector method that: \(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\).
Here,
D is the midpoint of the line BC of the \(\triangle\)ABC.
In \(\triangle\)ABD,
\(\vec {AB}\) = \(\vec {AD}\) + \(\vec {DB}\)....................(1) [\(\because\) triangle law of vector addition]
In \(\triangle\)ACD,
\(\vec {AC}\) = \(\vec {AD}\) + \(\vec {DC}\)...................(2)[\(\because\) triangle law of vector addition]
Adding (1) and (2)
\(\vec {AB}\) + \(\vec {AC}\) = \(\vec {AD}\) + \(\vec {BD}\) + \(\vec {AD}\) + \(\vec {DC}\)
or,\(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) + (\(\vec {DB}\) + \(\vec {DC}\))
or,\(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) + \(\vec {DB}\) - \(\vec {DB}\) [\(\because\) -\(\vec {DB}\) = \(\vec {DC}\)]
∴ \(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) Ans
The position vector of the point A and B are 3 + 2 and 5 - 6 respectively. Find the position vectors of the midpoint M of the line AB.
Let, O be the origin.
\(\vec {OA}\) = 3\(\vec i\) + 2\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 6\(\vec j\)
We know,
\(\vec {OM}\) = \(\frac {\vec {OA} + \vec {OB}}2\)
or, \(\vec {OM}\) = \(\frac {3\vec i + 2\vec j + 5\vec i - 6\vec j}2\)
or, \(\vec {OM}\) = \(\frac {8\vec i - 4\vec j}2\)
∴ \(\vec {OM}\) = 4\(\vec i\) - 2\(\vec j\)
∴ The position vector of M = 4\(\vec i\) - 2\(\vec j\) Ans
The position vectors of the points A and B are 5 + 2 and 3 + 6 respectively. Find the position vector of the point P which divides AB internally in the ratio 2 : 3.
Let: O be the origin.
\(\vec {OA}\) = 5\(\vec i\) + 2\(\vec j\) and \(\vec {OB}\) = 3\(\vec i\) + 6\(\vec j\)
m : n = 2 : 3
Using section internal division formula.
\(\vec {OP}\) = \(\frac {m \vec{OB} + n\vec {OA}}{m + n}\)
or, \(\vec {OP}\) = \(\frac {2(3\vec i + 6\vec j) + 3(5\vec i + 2\vec j)}{2 + 3}\)
or, \(\vec {OP}\) = \(\frac {6\vec i + 12\vec j + 15\vec i + 6\vec j}5\)
or, \(\vec {OP}\) = \(\frac {21\vec i + 18\vec j}5\)
∴ \(\vec {OP}\) = \(\frac {21}5\)\(\vec i\) + \(\frac {18}5\)\(\vec j\) Ans
If the co-ordinates of the points A and B are (-4, 8) and (3, 7) respectively. Find the position vector of the point which divides the line segment AB externally in the ratio 4 : 3.
Here,
A(-4, 8) and B(3, 7)
Let, O be the origin.
\(\vec {OA}\) = \(\begin {pmatrix} -4\\ 8\\ \end {pmatrix}\)
\(\vec {OB}\) = \(\begin {pmatrix} 3\\ 7\\ \end {pmatrix}\)
Using section external division formula,
\(\vec {AB}\) = \(\frac {4\begin {pmatrix} 3\\ 7\\ \end {pmatrix} - 3\begin {pmatrix} -4\\ 8\\ \end {pmatrix}}{4 - 3}\)
or, \(\vec {AB}\) = \(\frac {\begin {pmatrix} 12\\ 28\\ \end {pmatrix} - \begin {pmatrix} -12\\ 24\\ \end {pmatrix}}1\)
or, \(\vec {AB}\) = \(\begin {pmatrix} 12 + 12\\ 28 - 24\\ \end {pmatrix}\)
or, \(\vec {AB}\) = \(\begin {pmatrix} 24\\ 4\\ \end {pmatrix}\)
∴ \(\vec {AB}\) = 24\(\vec i\) + 4\(\vec j\) Ans
If \(\vec {OA}\) = \(\begin {pmatrix} \sqrt 3\\ 1\\ \end {pmatrix}\) and \(\vec {OB}\) = \(\begin {pmatrix} \sqrt 3\\ 3\sqrt 3\\ \end {pmatrix}\), calculate the \(\angle\)AOB.
Here,
\(\vec {OA}\) = \(\begin {pmatrix} \sqrt 3\\ 1\\ \end {pmatrix}\) and \(\vec {OB}\) = \(\begin {pmatrix} \sqrt 3\\ 3\sqrt 3\\ \end {pmatrix}\)
Let: x1 = \(\sqrt 3\), y1 = 1, x2 = \(\sqrt 3\) and y2 = 3\(\sqrt 3\)
Let: \(\theta\) be the angle between \(\vec {OA}\) and \(\vec {OB}\)
cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}}\)
or, cos \(\angle\)AOB = \(\frac {\sqrt 3 × \sqrt 3 + 1 × 3\sqrt 3}{\sqrt {(\sqrt 3)^2 + 1^2} \sqrt {(\sqrt 3)^2 + (3\sqrt 3)^2}}\)
or, cos \(\angle\)AOB = \(\frac {3 + 3\sqrt 3}{\sqrt {3 + 1}. \sqrt {3 + 27}}\)
or, cos \(\angle\)AOB = \(\frac {3 + 3\sqrt 3}{2\sqrt {30}}\)
or, cos \(\angle\)AOB = \(\frac {8.197}{10.95}\)
or, cos \(\angle\)AOB = 0.75
or, \(\angle\)AOB = cos-1 (0.75)
∴\(\angle\)AOB = 41.58° Ans
State parallelogram law of vector addition.
If the vector is \(\vec u\) and \(\vec v\) are represented by the two adjacent sides \(\vec {AB}\) and \(\vec {AD}\) of a parallelogram ABCD, then the sum \(\vec u\) + \(\vec v\) is represented by the diagonal \(\vec {AC}\).
\(\vec {AB}\) + \(\vec {AD}\) = \(\vec {AC}\)
\(\vec u\) + \(\vec v\) = \(\vec {AC}\)
Define unit vector and prove that: \(\vec a\) = 4\(\vec i\) - 6\(\vec j\) and \(\vec b\) = 3\(\vec i\) + 2\(\vec j\) are perpendicular.
Unit vector: A vector whose modulus is 1 is called a unit vector.
\(\vec a\) = (\(\frac 1{\sqrt 2}\), \(\frac 1{\sqrt 2}\)) is unit vector.
\(\vec a\) . \(\vec b\)
= (4\(\vec i\) - 6\(\vec j\)) . (3\(\vec i\) + 2\(\vec j\))
= 12\(\vec {i^2}\) + 8\(\vec i\) \(\vec j\) - 18\(\vec i\) \(\vec j\) - 12\(\vec {j^2}\)
= 12× 1 + 8× 0 - 18× 0 - 12× 1 [\(\vec {i^2}\) = \(\vec {j^2}\) = 1, \(\vec i\) . \(\vec j\) = 0]
= 12 - 12
= 0
\(\vec a\) . \(\vec b\) = 0
∴ \(\vec a\) . \(\vec b\) are perpendicular each other. Proved
If \(\vec a\) = (4\(\vec i\) + 5\(\vec j\)) and \(\vec b\) = (5\(\vec i\) - 4\(\vec j\)), find the angle between \(\vec a\) and \(\vec b\).
Here,
\(\vec a\) = (4\(\vec i\) + 5\(\vec j\)) and \(\vec b\) = (5\(\vec i\) - 4\(\vec j\))
\(\vec a\) = (4, 5) and \(\vec b\) = (5, -4)
If \(\theta\) be the angle between \(\vec a\) and \(\vec b\) then:
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)
or, cos\(\theta\) = \(\frac {(4, 5) . (5, -4)}{\sqrt {4^2 + 5^2} \sqrt {5^2 + (-4)^2}}\)
or, cos\(\theta\) = \(\frac {20 - 20}{\sqrt {16 + 25} \sqrt {25 + 16}}\)
or, cos\(\theta\) = \(\frac 0{25 + 16}\)
or, cos\(\theta\) = 0
or, \(\theta\) = cos-1 (0)
∴ \(\theta\) = 90° Ans
If \(\vec a\) = 10\(\vec i\) - 7\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 10\(\vec j\). Find the angle between \(\vec a\) and \(\vec b\).
Here,
\(\vec a\) = 10\(\vec i\) - 7\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 10\(\vec j\)
If \(\theta\) be the angle between the vectors \(\vec a\) and \(\vec b\).
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)
or, cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}}\)
or, cos\(\theta\) = \(\frac {10 × 7 + 10 × (-7)}{\sqrt {(10)^2 + (-7)^2} \sqrt {7^2 + {10}^2}}\)
or, cos\(\theta\) = \(\frac {70 - 70}{\sqrt {149} \sqrt {149}}\)
or, cos\(\theta\) = 0°
∴ \(\theta\) = cos-1(0°) = 90° Ans
If the position vectors of the vertices of \(\triangle\)ABC are A(-1, -1), B(5, -1) and C(2, 5). Find the position vector of the centroid G of the \(\triangle\)ABC.
In the \(\triangle\)ABC,
Let O be the origin.
Position vector of the point A is = \(\vec {OA}\) = (-1, -1)
Position vector of the point B is = \(\vec {OB}\) = (5, -1)
Position vector of the point C is = \(\vec {OC}\) = (2, 5)
If G be the centroid of the triangle,
= \(\frac 13\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\))
= \(\frac 13\) (- 1 - 1 + 5 - 1 + 2 + 5)
= \(\frac 13\) (6 + 3)
= 2 + 1
Hence, the position vector of the centroid is (2, 1). Ans
If \(\vec a\) = 6\(\vec i\) - 8\(\vec j\) and \(\vec b\) = 4\(\vec i\) + 3\(\vec j\) prove that the vector \(\vec a\) and \(\vec b\) are perpendicular to each other.
Here,
\(\vec a\) = 6\(\vec i\) - 8\(\vec j\) and \(\vec b\) = 4\(\vec i\) + 3\(\vec j\)
\(\vec a\) . \(\vec b\)
= (6\(\vec i\) - 8\(\vec j\)) (4\(\vec i\) + 3\(\vec j\))
= 6 . 4\(\vec {i^2}\) + 18\(\vec i\) \(\vec j\) - 32 \(\vec i\) \(\vec j\) - 24\(\vec {j^2}\)
= 24 . 1 + 18 . 0 - 32 . 0 * 24 . 1 [\(\because\) \(\vec {i^2}\) = \(\vec {j^2}\) = 1 , \(\vec j\) = \(\vec i\) \(\vec j\) = 0]
= 24 - 24
= 0
∴ The dot product of two vector a and b is equal to zero so these are perpendicular to each other. Proved
If \(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\), find the value of \(\angle\)AOB.
Here,
\(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\)
\(\vec a\) . \(\vec b\) = x1x2 + y1y2 = 7× 5 + (-5) (-7) = 35 + 35 = 70
\(\begin {vmatrix} \vec a \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {7^2 + (-5)^2}\) = \(\sqrt {49 + 25}\) = \(\sqrt {74}\)
\(\begin {vmatrix} \vec b\end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {5^2 + (-7)^2}\) = \(\sqrt {25 + 49}\) = \(\sqrt {74}\)
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a \end {vmatrix} \begin {vmatrix} \vec b\end {vmatrix}}\)
or, cos\(\theta\) = \(\frac {70}{\sqrt {74} . \sqrt {74}}\)
or, cos\(\theta\) = \(\frac {70}{74}\)
or, cos\(\theta\) = 0.95
or, \(\theta\) = cos-1 (0.95)
∴ \(\theta\) = 18.93°
∴ \(\angle\)AOB = 18.93° Ans
What do you mean by a unit vector? If the position vectors of A and B are 3\(\vec i\) + 4\(\vec j\) and 7\(\vec i\) + 8\(\vec j\) respectively, find the position vector of the mid-point of the line joining A and B.
Unit Vector: If the magnitude of the vector is 1 such type of vector is known as unit vector.
i.e., \(\vec {OP}\) = \(\begin {pmatrix} 1\\ 0\\ \end {pmatrix}\)
Let, \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 8\(\vec j\) are the position vectors of A and B,
Position vector of mid-point \(\vec M\) = \(\frac 12\) (\(\vec a\) + \(\vec b\))
\(\vec M\) = \(\frac 12\) (3\(\vec i\) + 4\(\vec j\) + 7\(\vec i\) + 8\(\vec j\)) = \(\frac 12\) (10\(\vec i\) + 12\(\vec j\)) = 5\(\vec i\) + 6\(\vec j\) Ans
In the given figure, \(\vec {OA}\) = \(\vec a\) and \(\vec {OB}\) = \(\vec b\). If \(\vec {AC}\) = 3\(\vec {AB}\), find \(\vec {OC}\).
Here,
\(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and\(\vec {AC}\) = 3\(\vec {AB}\) then, \(\vec {OC}\) = ?
Using triangle law of vector addition:
\(\vec {OA}\) + \(\vec {AB}\) \(\vec {OB}\)
or, \(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) = \(\vec b\) - \(\vec a\)
We know,
\(\vec {AC}\) = 3\(\vec {AB}\) = 3(\(\vec b\) - \(\vec a\)) = 3\(\vec b\) - 3\(\vec a\)
Again,
\(\vec {OA}\) + \(\vec {AC}\) = \(\vec {OC}\)
or, \(\vec {OC}\) = \(\vec a\) + 3\(\vec b\) - 3\(\vec a\)
∴ \(\vec {OC}\) = 3\(\vec b\) - 2\(\vec a\) Ans
On what condition \(\vec a\) and \(\vec b\) are parallel? If \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 6\(\vec j\), find the angle between the vectors \(\vec a\) and \(\vec b\).
The two vectors are \(\vec a\) and \(\vec b\) are parallel if \(\vec a\) = m\(\vec b\) where m is scalar product.
\(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 6\(\vec j\)
\(\vec a\).\(\vec b\) = x1x2 + y1y2 = 3× 8 + (-4) × 6= 24 - 24 = 0
\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5
\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-6)^2}\) = \(\sqrt {64 + 36}\) = \(\sqrt {100}\) = 10
We know,
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac 0{5 × 10}\) = 0
∴ \(\theta\) = cos-1 (0) = 90° Ans
Point C divides the line AB internally in the ratio of 3 : 1. If the position vectors of A and B are \(\vec i\) - 3\(\vec j\) and 2\(\vec i\) + 5\(\vec j\) respectively, find \(\begin {vmatrix} \vec {AB} \end {vmatrix}\) and the position vector of point C.
Let: \(\vec a\) = \(\vec i\) - 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) - 5\(\vec j\) , m:n = 3:1
Position vector of C (\(\vec {OC}\)) = \(\frac {n\vec a + m \vec b}{m + n}\)
\(\vec c\) = \(\vec {1 (\vec i - 3\vec j)}{3 + 1}\)
\(\vec c\) = \(\frac {\vec i - 3\vec j + 6\vec i + 15\vec j}{4}\)
\(\vec c\) = \(\frac {7\vec i + 12\vec j}4\) Ans
\(\vec {AB}\) = \(\begin {pmatrix} x_2 - x_1\\ y_2 - y_1\\ \end {pmatrix}\) =\(\begin {pmatrix} 2 - 1\\ 5 + 3\\ \end {pmatrix}\) = \(\begin {pmatrix} 1\\ 8\\ \end {pmatrix}\)
\(\begin {vmatrix} \vec {AB}\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {1^2 + 8^2}\) = \(\sqrt {1 + 64}\) = \(\sqrt {65}\) Ans
In the figure, \(\vec {OA}\) = \(\vec a\) . \(\vec {OB}\) = \(\vec b\) and \(\vec {AC}\) = 5\(\vec AB\), find \(\vec {OC}\) in terms of \(\vec a\) and \(\vec b\).
Here,
Using triangle law of vector addition,
\(\vec {OC}\) = \(\vec {OA}\) + \(\vec {AC}\)
or, \(\vec {OC}\) = \(\vec a\) + 5\(\vec {AB}\)
or, \(\vec {OC}\) = \(\vec a\) + 5 (\(\vec {OB}\) - \(\vec {OA}\))
or, \(\vec {OC}\) = \(\vec a\) + 5 (\(\vec b\) - \(\vec a\))
or, \(\vec {OC}\) = \(\vec a\) + 5\(\vec b\) - 5\(\vec a\)
∴ \(\vec {OC}\) = 5\(\vec b\) - 4\(\vec a\) Ans
What do you mean by orthogonal vectors? If \(\vec p\) = \(\begin {pmatrix} 3\\ -2\\ \end {pmatrix}\) and \(\vec q\) = \(\begin {pmatrix} 2\\ 3\\ \end {pmatrix}\), show that \(\vec p\) and \(\vec q\) are orthogonal vectors.
Orthogonal Vectors: If two vectors are perpendicular each other then vector are known as orthogonal vector.
\(\vec p\) . \(\vec q\) = 3× 2 - 2× 3 = 0
∴ The dot product od \(\vec p\) and \(\vec q\) is equal to zero, so these vectors are orthogonal. Proved
If the position vectors of the points A and B are 3\(\vec i\) - 2\(\vec j\) and 1\(\vec i\) + 8\(\vec j\) respectively, find the position vector of the mid-point of the line AB.
Let: \(\vec {OA}\) = 3\(\vec i\) - 2\(\vec j\) and \(\vec {OB}\) = 1\(\vec i\) + 8\(\vec j\)
Mid-point of \(\vec {AB}\)
= \(\frac {\vec {OA} + \vec {OB}}2\)
= \(\frac {3\vec i - 2\vec j + 1\vec i + 8\vec j}2\)
= \(\frac {4\vec i + 6\vec j}2\)
= 2\(\vec i\) + 3\(\vec j\)
∴ Position vector of the mid-point of \(\vec {AB}\) = 2\(\vec i\) + 3\(\vec j\) Ans
Define position vector. If P is the mid-point of the straight line joining the points A(3, 4) and B(-1, 2), find the position vector of P.
Position Vectors: Let P(x, y) be any point on the co-ordinates axes, then (x, y) is called the position vector of the point P with referred to the origin O. OP is a position vector.
Let: \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = -\(\vec i\) + 2\(\vec j\)
\(\vec m\) = ?
\(\vec m\) = \(\frac {\vec a + \vec b}2\) = \(\frac {3\vec i + 4\vec j - \vec i + 2\vec j}2\) = \(\frac {2\vec i + 6\vec j}2\) = \(\vec i\) + 3\(\vec j\) Ans
If PQRS is a quadrilateral, prove that:
\(\vec {QR}\) + \(\vec {RS}\) + \(\vec {SP}\) = \(\vec {QP}\)
Using triangle law of vector addition,
In \(\triangle\)QRS,
\(\vec {QS}\) = \(\vec {QR}\) + \(\vec {RS}\) ....................................(1)
In \(\triangle\)PQS,
\(\vec {QS}\) = \(\vec {QP}\) + \(\vec {PS}\) ...................................(2)
From (1) and (2),
\(\vec {QR}\) + \(\vec {RS}\) = \(\vec {QP}\) + \(\vec {PS}\)
or, \(\vec {QR}\) + \(\vec {RS}\) - \(\vec {PS}\) = \(\vec {QP}\)
∴ \(\vec {QR}\) + \(\vec {RS}\) +\(\vec {SP}\) = \(\vec {QP}\) Proved
If \(\vec a\) = \(\vec i\) + 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) + \(\vec j\), find the angle between \(\vec a\) and \(\vec b\).
Given,
\(\vec a\) = \(\vec i\) + 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) + \(\vec j\)
Let an anglebetween \(\vec a\) and \(\vec b\) be \(\theta\)
So,
cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a \end {vmatrix} \begin {vmatrix} \vec b \end {vmatrix}}\)
or, cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {{x_1}^2 +{y_1}^2} \sqrt {{x_2}^2 + {y_2}^2}}\) [where x1 = 1, y1 = 3, x2 = 2 and y2 = 1]
or, cos\(\theta\) = \(\frac {1 × 2 + 3 × 1}{\sqrt {1^2 + 3^2} \sqrt {2^2 + 1^2}}\)
or, cos\(\theta\) = \(\frac {2 + 3}{\sqrt {10} \sqrt 5}\)
or, cos\(\theta\) = \(\frac 5{\sqrt {50}}\)
or, cos\(\theta\) = \(\frac 5{5\sqrt 2}\)
or, cos\(\theta\) = \(\frac 1{\sqrt 2}\)
or, cos\(\theta\) = cos 45°
∴ \(\theta\) = 45° Ans
define a null vector. If \(\vec a\) = \(\begin {pmatrix} -5\\ 3\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} p\\ p + 2\\ \end {pmatrix}\) are perpendicular to each other, find the value of p.
Null Vector: If the magnitude of vector is equal to zero such type of vector is known as zero or null vector. \(\vec {AA}\) = 0
If \(\vec a\) and \(\vec b\) are perpendicular then,
\(\vec a\) . \(\vec b) = 0
or, (-5, 3) (p, p+2) = 0
or, -5× p + 3× (p + 2) = 0
or, -5p + 3p + 6 = 0
or, -2p = - 6
or, p = \(\frac 62\)
∴ p = 3 Ans
Prove that: \(\vec {PQ}\) + \(\vec {QR}\) + \(\vec {RS}\) + \(\vec {SP}\) = 0 in the given quadrilateral PQRS.
Joining point P and R,
In \(\triangle\)RSP,
\(\vec {RP}\) = \(\vec {RS}\) + \(\vec {SP}\).................................(1)
In \(\triangle\)PQR,
\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)..............................(2)
Adding equation (1) and (2),
\(\vec {RP}\) + \(\vec {PR}\) = \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\)
or, -\(\vec {PR}\)+ \(\vec {PR}\) = \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\) [\(\because\) \(\vec {RP}\) = -\(\vec {PR}\)]
or, 0= \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\)
∴ \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\) = 0 Proved
Prove that: \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) = 0 in given triangle ABC.
\(\vec {AB}\), \(\vec {BC}\) and \(\vec {AC}\) are the vertices which represent the side of \(\triangle\)ABC using triangle law of addition.
\(\vec {BC}\) + \(\vec {CA}\) = \(\vec {BA}\)
or, \(\vec {BC}\) + \(\vec {CA}\) = -\(\vec {AB}\) [\(\because\) \(\vec {BA}\) = - \(\vec {AB}\)]
∴ \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) = 0 Proved
In the given diagram, prove that:
\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) + \(\vec {EA}\) = 0
In \(\triangle\)ABC,
\(\vec {AB}\) + \(\vec {BC}\) = \(\vec {AC}\)............................(1)
In \(\triangle\)ACD,
\(\vec {AC}\) + \(\vec {CD}\) = \(\vec {AD}\)...........................(2)
In \(\triangle\)ADE,
\(\vec {AD}\) + \(\vec {DE}\) = \(\vec {AE}\).............................(3)
or, \(\vec {AC}\) + \(\vec {CD}\) + \(\vec {DE}\) = \(\vec {AE}\)
or, \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) = \(\vec {AE}\)
∴\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) + \(\vec {EA}\) = 0 Proved
In the given figure, ABCD is a parallelogram. If \(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\), \(\vec {OC}\) = \(\vec c\), find \(\vec {OD}\).
Here,
Using triangle law of vector addition,
In \(\triangle\)OBC,
\(\vec {BC}\) = \(\vec {BO}\) + \(\vec {OC}\)
∴ \(\vec {BC}\) = -\(\vec b\) + \(\vec c\)
In \(\triangle\)OAD,
\(\vec {OD}\) = \(\vec {OA}\) + \(\vec {AD}\) = \(\vec {OA}\) + \(\vec {BC}\) [\(\vec {AD}\) = \(\vec {BC}\)]
∴ \(\vec {OD}\) = \(\vec a\) - \(\vec b\) + \(\vec c\) Ans
Prove by a vector method that the parallelogram with equal diagonals is a rectangle.
Let, PQRS be a parallelogram which PR and SQ are equal diagonals.
Let: \(\vec {SR}\) = \(\vec a\) and \(\vec {SP}\) = \(\vec b\)
To prove: PQRS is a parallelogram.
\(\vec {SQ}\) = \(\vec {SR}\) + \(\vec {RQ}\) [\(\because\) Triangle law of vector addition]
\(vec {SR}\) = \(\vec {SR} \) + \(\vec {SP}\) [\(\because\) \(\vec {RQ}\) = \(\vec {SP}\)]
\(\vec {SQ}\) = \(\vec a\) + \(\vec b\)
\(\vec {RP}\) =\(\vec {RS}\) + \(\vec {SP}\) = -\(\vec {SR}\) + \(\vec {SP}\) = -\(\vec a\) \(\vec b\) = \(\vec b\) - \(\vec a\)
\(\begin {vmatrix} \vec {SQ}\\ \end {vmatrix}\) = \(\begin {vmatrix} \vec {RP}\\ \end {vmatrix}\)
or, \(\vec {SQ}\)2 = \(\vec {RP}\)2
or, (\(\vec a\) + \(\vec b\))2 = (\(\vec b\) - \(\vec a\))2
or, \(\vec a\)2 + 2\(\vec a\) \(\vec b\) + \(\vec b\)2 = \(\vec b\)2- 2\(\vec a\) \(\vec b\) + \(\vec a\)2
or, 4\(\vec a\) \(\vec b\) = 0
or, \(\vec a\) . \(\vec b\) = 0
∴ \(\angle\)PSR = 90°
Hence, PQRS is a rectangle. Proved
Prove by vector method that diagonals of a rhombus bisect each at right angle.
Let: ABCD is a rhombus.
Let: \(\vec {AB}\) = \(\vec a\) and \(\vec {AD}\) \ \(\vec b\)
To prove: AC ⊥BD and AO = \(\frac 12\) AC, BO = \(\frac {BD}\).
In \(\triangle\)ABC,
\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {BC}\) [\(\because\) Triangle law of vector addition]
\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {AD}\) [\(\because\) BC = AD]
\(\vec {AC}\) = \(\vec a\) + \(\vec b\)
\(\vec {BD}\) = \(\vec {BA}\) + \(\vec {AD}\) = -\(\vec a\) + \(\vec b\) = \(\vec b\) - \(\vec a\)
\(\vec {AC}\) . \(\vec {BD}\)
= (\(\vec a\) + \(\vec b\)) (\(\vec b\) - \(\vec a\))
= \(\vec b\)2 - \(\vec a\)2
=\(\vec {AD}\)2 - \(\vec {AB}\)2
= \(\vec {AB}\)2 - \(\vec {AB}\)2 [\(\because\) AB = AD]
= 0
∴ AC ⊥ BD
\(\vec {AO}\)= \(\vec {AB}\) + \(\vec {BO}\) = \(\vec {AB}\) + \(\frac 12\) \(\vec {BD}\) = a + \(\frac 12\)(\(\vec b\) - \(\vec a\))
\(\vec {AO}\) = \(\frac {2a + \vec b - \vec a}2\) = \(\frac 12\)(\(\vec a\) + \(\vec b\)) = \(\frac 12\)\(\vec {AC}\)..........................(1)
\(\vec {BO}\) = \(\vec {BA}\) + \(\vec {AO}\) = -\(\vec {AB}\) + \(\vec {AO}\) = - \(\vec a\) + \(\frac 12\)(\(\vec a\) + \(\vec b\)
\(\vec {BO}\) = \(\frac {-2\vec a + \vec a + \vec b}2\) = \(\frac 12\)(\(\vec b\) - \(\vec a\)) = \(\frac 12\)\(\vec {BD}\)............(2)
From relation (1) and (2):
Diagonal bisect each other.
Hence, diagonals of a rhombus bisects each at right angle. Proved
Prove by vector method that opposite sides of a parallelogram are equal.
Let: PQRS is a parallelogram, in which PQ//SR and PS//QR.
To prove: PQ = SR and PS = QR
Let \(\vec {PQ}\) = m\(\vec {SR}\) and \(\vec {PS}\) = n\(\vec {QR}\) where m and n are scalars.
In \(\triangle\)PQR,
\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)
\(\vec {PR}\) = m\(\vec {SR}\) + n\(\vec {QR}\)......................(1)
In \(\triangle\)PSR,
\(\vec {PR}\) = \(\vec {PS}\) + \(\vec {SR}\)...............................(2)
From (1) and (2),
m\(\vec {SR}\) + n\(\vec {QR}\) = \(\vec {PS}\) + \(\vec {SR}\)
or, m\(\vec {SR}\) - \(\vec {SR}\) = \(\vec {QR}\) - n\(\vec {QR}\)
or, \(\vec {SR}\) (m - 1) = \(\vec {QR}\) (1 - n)
\(\vec {QR}\) and \(\vec {SR}\) are not parallel so;
m - 1 = 0∴ m = 1
1 - n = 0∴ n = 1
Putting the value of m and n in
\(\vec {PQ}\) = m\(\vec {SR}\) and \(\vec {PS}\) = n\(\vec {QR}\)
\(\vec {PQ}\) = \(\vec {SR}\) and \(\vec {PS}\) = \(\vec {QR}\)
Hence, opposite sides of a parallelogram are equal. Proved
If diagonal of parallelogram are at right angles to each other prove by vector method that it is a rhombus.
ABCD is a parallelogram, diagonals AC and BD are perpendicular to each other.
To prove: ABCD is a rhombus.
Let: \(\vec {AB}\) = \(\vec {DC}\) = \(\vec a\) and \(\vec {BC}\) = \(\vec {AD}\) = \(\vec b\)
Using triangle law of vector addition:
In \(\triangle\)ABC,
\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {BC}\)
\(\vec {AC}\) = \(\vec a\) + \(\vec b\)
In \(\triangle\)BCD,
\(\vec {BD}\) = \(\vec {BC}\) + \(\vec {CD}\)
\(\vec {BD}\) = \(\vec {BC}\) - \(\vec {DC}\) = \(\vec b\) - \(\vec a\)
AC and BD are perpendicular to each other.
\(\vec {AC}\) . \(\vec {BD}\) = 0
(\(\vec a\) + \(\vec b\)) + (\(\vec b\) - \(\vec a\)) = 0
or, \(\vec {b^2}\) - \(\vec {a^2}\) = 0
or, \(\vec {b^2}\) = \(\vec {a^2}\)
or, \(\begin {vmatrix} \vec {b^2} \end {vmatrix}\) =\(\begin {vmatrix} \vec {a^2} \end {vmatrix}\)
∴\(\begin {vmatrix} \vec b\end {vmatrix}\) =\(\begin {vmatrix} \vec a\end {vmatrix}\)
AB = BC
Hence, ABCD is a rhombus. Proved
Prove by vector method that the sum of the squares of the diagonals of any parallelogram is equal to the sum of the squares of the four sides.
PQRS is a parallelogram, PR and QS are diagonals.
To prove: PR2+ QS2 = PQ2 + QR2 + RS2 + PS2
In \(\triangle\)PQR,
\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\) [\(\because\) Triangle law of vector addition]
Squaring on both sides,
\(\vec {PR}^2\) = (\(\vec {PQ}\) + \(\vec {QR}\))2
\(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 . \(\vec {PQ}\) . \(\vec {QR}\) + \(\vec {QR}^2\)
PR2 = PQ2 + QR2 + 2 PQ. QR.....................(1)
In \(\triangle\)QRS,
\(\vec {QS}\) = \(\vec {QR}\) + \(\vec {RS}\) [\(\because\) Triangle law of vector addition]
Squaring on both sides,
\(\vec {QS}^2\) = (\(\vec {QR}\) + \(\vec {RS}\))2 = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . \(\vec {RS}\) + \(\vec {RS}^2\)
\(\vec {QS}^2\) = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . \(\vec {QP}\) + \(\vec {RS}^2\) [\(\because\) \(\vec {RS}\) = \(\vec {QP}\)]
\(\vec {QS}^2\) = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . (-\(\vec {PQ}\)) + \(\vec {RS}^2\)
\(\vec {QS}^2\) = \(\vec {QR}^2\) -2 . \(\vec {QR}\) . \(\vec {PQ}\) + \(\vec {RS}^2\)
QS2 = QR2 + RS2 - 2 . PQ . QR............................(2)
Adding (1) and (2)
PR2 + QS2 = PQ2 + QR2 + QR2 + RS2 + 2 PQ . QR - 2 PQ . QR
∴PR2 + QS2 = PQ2 + QR2 + PS2 + RS2Hence, Proved
In the given figure, PQRS is a trapezium where PQ\\SR. A and B are the mid-points of PR and QS respectively. Prove by vector method that
PQRS is a trapezium in which PQ\\SR. A and B are the mid-points of PR and QS. Let, P be the origin.
To prove: i. AB//PQ//SR ii. AB = \(\frac 12\)(PQ - SR)
Construction: join PB
Let, P be the origin.
\(\vec {SR}\) = m\(\vec {PQ}\) [\(\because\) PQ//SR]
\(\vec {PR}\) = \(\vec {PS}\) + \(\vec {SR}\) [Using triangle law of vector addition]
\(\vec {PR}\) = \(\vec {PS}\) + m\(\vec {PQ}\)
\(\vec {PB}\) = \(\frac {\vec {PS} + \vec {PQ}}2\) [\(\because\) B is the mid-point of QS]
\(\vec {PA}\) = \(\frac 12\)\(\vec {PR}\) = \(\frac 12\)(\(\vec {PS}\) + m\(\vec {PQ}\))
\(\vec {AB}\) = \(\vec {PB}\) - \(\vec {PA}\) = \(\frac 12\)(1 - m)\(\vec {PQ}\)
\(\vec {AB}\)//\(\vec {PQ}\)
∴ AB//PQ//SR
Again,
\(\vec {AB}\) = \(\frac 12\)\(\vec {PQ}\) - \(\frac 12\)m\(\vec {PQ}\) = \(\frac 12\)(\(\vec {PQ}\) - m\(\vec {PQ}\)) = \(\frac 12\)(\(\vec {PQ}\) - \(\vec {SR}\))
∴ AB = \(\frac 12\)(PQ - SR) Proved
In the given figure, PQRS is a parallelogram. M and N are two points on the diagonal SQ. If SM = NQ, prove by vector method that PMRN is a parallelogram.
Given:
PQRS is a parallelogram.
In which: PQ =SR, PQ//SR and PS = QR, PS//QR
To prove: PMRN is a parallelogram.
In \(\triangle\)PQN,
\(\vec {PQ}\) = \(\vec {PN}\) + \(\vec {NQ}\)....................(1) [Using triangle law of vector addition]
In \(\triangle\)SMR,
\(\vec {SR}\) = \(\vec {SM}\) + \(\vec {MR}\).....................(2)
From equation (1) and (2)
\(\vec {PQ}\) = \(\vec {SR}\)
or, \(\vec {PN}\) + \(\vec {NQ}\) = \(\vec {SM}\) + \(\vec {MR}\)
\(\vec {PN}\) = \(\vec {MR}\) [\(\because\) \(\vec {NQ}\) = \(\vec {SM}\)]
PN and MR are equal and parallel.
Similarly, PM and NR are equal and parallel.
∴ PMRN is a parallelogram. Proved
If D, E and F are the mid-point of the sides of the triangle ABC respectively and O is any point, prove that:
\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) = \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\)
In the figure, ABC is a triangle with medians. AE, BF and CD of sides BC, AC and AB respectively. BE = \(\frac 12\)BC, CF = \(\frac 12\)CA, AD = \(\frac 12\)AB.
To prove:\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) = \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\)
Using triangle law of vector addition,
\(\vec {AE}\) = \(\vec {AB}\) + \(\vec {BE}\) = \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\)........................(1)
\(\vec {BF}\) = \(\vec {BC}\) + \(\vec {CF}\) = \(\vec {BC}\) + \(\frac 12\)\(\vec {CA}\)....................(2)
\(\vec {CD}\) = \(\vec {CA}\) + \(\vec {AD}\) = \(\vec {CA}\) + \(\frac 12\)\(\vec {AB}\)......................(3)
Adding (1), (2) and (3),
\(\vec {AE}\) + \(\vec {BF}\) + \(\vec {CD}\)
= \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\) +\(\vec {BC}\) + \(\frac 12\)\(\vec {CA}\) +\(\vec {CA}\) + \(\frac 12\)\(\vec {AB}\)
= \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) + \(\frac 12\)(\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\))
= \(\vec {AC}\) + \(\vec {CA}\) + \(\frac 12\)(\(\vec {AC}\) + \(\vec {CA}\)) [\(\because\) \(\vec {AB}\) + \(\vec{BC}\) = \(\vec {AC}\)]
\(\vec {AE}\) + \(\vec {BF}\) + \(\vec {CD}\) = -\(\vec {CA}\) + \(\vec {CA}\) + \(\frac 12\)(-\(\vec {CA}\) + \(\vec {CA}\)) = 0
or, \(\vec {AO}\) + \(\vec {OE}\) + \(\vec {BO}\) + \(\vec {OF}\) + \(\vec {CO}\) + \(\vec {OD}\) = 0 [\(\because\) \(\vec {AE}\) = \(\vec {AO}\) + \(\vec {OE}\)]
or, \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) - \(\vec {OA}\) - \(\vec {OB}\) - \(\vec {OC}\) = 0
or, \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) = \(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\)
∴\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) =\(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) Hence, Proved
In the given trapezium ABCD, AD//BC. M and N are the mid-point of AB and DC respectively. Prove by vector method:
MN = \(\frac 12\)(AD + BC)
Given:
ABCD is a trapezium.
AD//BC//MN and M and N are the mid-point of AB and DC.
To prove:MN = \(\frac 12\)(AD + BC)
Construction: join point A and N, join point B and N
Using triangle law of vector addition:
In \(\triangle\)BMN,
\(\vec {MN}\) = \(\vec {MB}\) + \(\vec {BN}\) = \(\vec {MB}\) + (\(\vec {BC}\) + \(\vec {CN}\))........................(1)
In \(\triangle\)MAN,
\(\vec {MN}\) = \(\vec {MA}\) + \(\vec {AN}\) = \(\vec {MB}\) + (\(\vec {AD}\) + \(\vec {DN}\))........................(2)
Adding (1) and (2)
2\(\vec {MN}\) = \(\vec {MB}\) + \(\vec {BC}\) + \(\vec {CN}\) + \(\vec {MA}\) + \(\vec {AD}\) + \(\vec {DN}\)
or, 2\(\vec {MN}\) = (\(\vec {MB}\) + \(\vec {BM}\)) + \(\vec {BC}\) + \(\vec {AD}\) + \(\vec {CN}\) + \(\vec {DN}\)
or, 2\(\vec {MN}\) = \(\vec {MB}\) - \(\vec {MB}\) + \(\vec {BC}\) + \(\vec {AD}\) + \(\vec {CN}\) + \(\vec {DN}\)
or, 2\(\vec {MN}\) = \(\vec {BC}\) + \(\vec {AD}\)
∴MN = \(\frac 12\)(AD + BC) Hence, Proved
Prove by vector method that a line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.
Given:
In \(\triangle\)ABC, M and N are the mid-point of Ab and AC. M and N are joined.
To prove: \(\vec {MN}\) //\(\vec {BC}\) and \(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\)
Using triangle law of vector addition,
In \(\triangle\)AMN,
\(\vec {MN}\) = \(\vec {MA}\) + \(\vec {AN}\).........................(1)
In \(\triangle\)ABC,
\(\vec {BC}\) = \(\vec {BA}\) + \(\vec {AC}\)...........................(2)
Since, M and N are mid-point of BA and AC.
\(\vec {BC}\) = 2\(\vec {MA}\) + 2\(\vec {AN}\)
\(\vec {BC}\) = 2 (\(\vec {MA}\) + \(\vec {AN}\))
\(\frac {\vec {BC}}2\) = \(\vec {MN}\) [From equation (1)]
\(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\)
\(\vec {MN}\)//\(\vec {BC}\) [\(\because\) \(\vec a\) = m\(\vec b\) where, m is a scalar quantity then \(\vec a\)//\(\vec b\)]
∴ \(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\) and \(\vec {MN}\)//\(\vec {BC}\) Hence, Proved
In the given figure, P and Q are the middle points of AB and AC respectively of the triangle ABC. Prove vectorially that PQ//BC and BC = 2PQ.
Given:
In \(\triangle\)ABC, P and Q are the mid points of AB and AC.
To Prove: \(\vec {PQ}\)// \(\vec {BC}\) and \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\)
Using triangle law of vector addition:
In \(\triangle\)APQ,
\(\vec {PQ}\) = \(\vec {PA}\) + \(\vec {AQ}\)....................................(1)
In \(\triangle\)ABC,
\(\vec {BC}\) = \(\vec {BA}\) + \(\vec {AC}\).....................................(2)
Dividing by 2 on both sides of equation (2)
\(\frac 12\)\(\vec {BC}\) =\(\frac 12\)(\(\vec {BA}\) + \(\vec {AC}\))
\(\frac 12\)\(\vec {BC}\) = \(\vec {PA}\) + \(\vec {AQ}\) [\(\because\) P and Q are the mid-point of AB and AC]
\(\frac 12\)\(\vec {BC}\) = \(\vec {PQ}\) [From equation (1)]
∴ \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\)
\(\vec {PQ}\)//\(\vec {BC}\) [\(\because\) \(\vec a\) = m\(\vec b\) then \(\vec a\)//\(\vec b\)]
Hence, \(\vec {PQ}\)//\(\vec {BC}\) and \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\) Hence, Proved
ABCD is a parallelogram and G is the point of intersection of its diagonals. If O is any point. Prove that \(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\) = 4\(\vec {OG}\).
AC and BD are the diagonals of a parallelogram of ABCD which intersect at a point G.GA = GC and GB = GD. If O be any point, then join O and A, O and B, O and C, O and D and O and G.
Using triangle law in vector addition,
In \(\triangle\)OAC,
\(\vec {OA}\) + \(\vec {OC}\) = \(\vec {AC}\) = 2\(\vec {OG}\).............................(1)
[\(\because\) G is a mid-point of AC]
In \(\triangle\)OBD,
\(\vec {OB}\) + \(\vec {OD}\) = 2\(\vec {OG}\)..........................................(2)
Adding equation (1) and (2)
\(\vec {OA}\) + \(\vec {OC}\) + \(\vec {OB}\) + \(\vec {OD}\) = 2\(\vec {OG}\) + 2\(\vec {OG}\)
∴\(\vec {OA}\) + \(\vec {OC}\) + \(\vec {OB}\) + \(\vec {OD}\) = 4\(\vec {OG}\) Hence, Proved
In the given figure, P, Q, R and S are the mid-points of the sides AB, AD, CD and BC respectively of the quadrilateral ABCD. Prove by vector method that PR and QS bisect each other.
Position vector of P, \(\vec {OP}\) = \(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\))
Position vector of R, \(\vec {OR}\) = \(\frac 12\)(\(\vec {OD}\) + \(\vec {OC}\))
Position vector of mid-point of \(\vec {PR}\)
= \(\frac 12\)[\(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\)) + \(\frac 12\)(\(\vec {OD}\) + \(\vec {OC}\))]
= \(\frac 12\) [\(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\))]
= \(\frac 14\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\))
Position vector of Q, \(\vec {OQ}\) = \(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\))
Position vector of S, \(\vec {OS}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OC}\))
Position vector of mid-point of \(\vec {QS}\)
= \(\frac 12\)[\(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\)) + \(\frac 12\)(\(\vec {OB}\) + \(\vec {OC}\))]
= \(\frac 12\) [\(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\) + \(\vec {OB}\) + \(\vec {OC}\))]
= \(\frac 14\)(\(\vec {OA}\) + \(\vec {OD}\) + \(\vec {OB}\) + \(\vec {OC}\))
∴ Position vector of mid-point of \(\vec {PR}\) and \(\vec {QS}\) is same. So, \(\vec {PR}\) and \(\vec {QS}\) bisect each other. Proved
Prove that the middle point of the hypotenuse of a right-angled triangle is equidistance from its vertices by vector method.
Let, \(\triangle\)ABC be the right angled triangle whose right angle is at A. Let, P be the middle point of the hypotenuse BC.
Take A is the origin and let;
\(\vec {AC}\) = \(\vec c\), \(\vec {AB}\) = \(\vec b\)
\(\vec {AP}\) = \(\vec d\) then \(\vec {AB}\) = \(\vec {AP}\) + \(\vec {PB}\)
[Using triangle law of vector addition]
\(\vec b\) = \(\vec d\) + \(\vec {PB}\) and \(\vec {AC}\) = \(\vec {AP}\) + \(\vec {PC}\)
or, \(\vec c\) = \(\vec d\) - \(\vec {CP}\) i.e. \(\vec c\) = \(\vec d\) - \(\vec {PB}\) [\(\because\) PB = PC]
Now,
It is given that: AB⊥ AC
∴ \(\vec b\) . \(\vec c\) = 0
or, (\(\vec d\) + \(\vec {PB}\)) (\(\vec d\) - \(\vec {PB}\)) = 0
or, d2 - PB2 = 0
or, d2 = PB2
∴ d = PB = PC = AP
∴ The middle point of the hypotenuse of a right-angled triangle is equidistance from the vertices. Proved
In the given figure, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively of the quadrilateral ABCD. Prove by vector method that PQRS is a parallelogram.
Let: ABCD be quadrilateral.
Take A as origin and suppose the position vectors of the vertices B, C, D are \(\vec a\), \(\vec b\) and \(\vec c\) respectively.
Let, P, Q, R and S be the mid-points od AB, BC, CD and DA. PQ, QR, RS and SP are joined.
By using the mid-point formula. We get the position vectors of P, Q, R and S as follows:
\(\vec {AP}\) = \(\frac {\vec o + \vec a}2\) = \(\frac 12\)\(\vec a\)
\(\vec {AQ}\) = \(\frac {\vec a + \vec b}2\) = \(\frac 12\)\(\vec a\) + \(\frac 12\)\(\vec b\)
\(\vec {AR}\) = \(\frac {\vec b + \vec c}2\) = \(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\)
\(\vec {AS}\) = \(\frac {\vec o + \vec c}2\) = \(\frac 12\)\(\vec c\)
\(\vec {QR}\)
= (position vector of R) - (position vector of Q)
= (\(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\)) - (\(\frac 12\)\(\vec a\) + \(\frac 12\) \(\vec b\))
= \(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\) - \(\frac 12\)\(\vec b\)
= \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\)..................................(1)
Also,
\(\vec {PS}\)
= (position vector of S) - (position vector of P)
= \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\)..................................(2)
From relation (1) and (2)
\(\vec {QR}\) = \(\vec {PS}\)
∴ QR = PS and QR//PS
∴ PQRS is a parallelogram. Proved
In the given figure, AD, BE and CF are the medians of the triangle ABC, prove that:
\(\vec {AD}\) + \(\vec {BE}\) + \(\vec {CF}\) = 0
D, E and F are the mid-points of BC, CA and AB.
By using triangle law of vector addition,
\(\vec {AB}\) + \(\vec {BC}\) = \(\vec {AC}\)
∴ \(\vec {BC}\) = \(\vec {AC}\) - \(\vec {AB}\)
Again,
\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {BD}\)..................................(1)
\(\vec {AD}\) = \(\vec {Ac}\) + \(\vec {CD}\)...................................(2)
Adding (1) and (2)
2\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {AC}\) + \(\vec {BD}\) + \(\vec {CD}\)
or, 2\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {AC}\) +0 [\(\because\) \(\vec {BD}\) = -\(\vec {CD}\)]
∴ \(\vec {AD}\) = \(\frac 12\)(\(\vec {AB}\) + \(\vec {AC}\))
\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\) = -\(\vec {AB}\) + \(\frac 12\)\(\vec {AC}\) = \(\frac 12\)\(\vec {AC}\) - \(\vec {AB}\)
\(\vec {CF}\) = \(\vec {CA}\) + \(\vec {AF}\) = -\(\vec {AC}\) + \(\frac 12\)\(\vec {AB}\) = \(\frac 12\)\(\vec {AB}\) - \(\vec {AC}\)
Now,,
\(\vec {AD}\) +\(\vec {BE}\) +\(\vec {CF}\)
=\(\frac 12\)(\(\vec {AB}\) + \(\vec {AC}\)) +\(\frac 12\)\(\vec {AC}\) - \(\vec {AB}\) +\(\frac 12\)\(\vec {AB}\) - \(\vec {AC}\)
= \(\vec {AB}\) + \(\vec {AC}\) - \(\vec {Ab}\) - \(\vec {AC}\)
= 0
∴\(\vec {AD}\) + \(\vec {BE}\) + \(\vec {CF}\) = 0 Hence, Proved
In the following figure, M and N are the mid-points of AB and BC respectively. Use vector method to prove that: \(\vec {AC}\) = 2\(\vec {MN}\).
Let: the position vector of A, B and C are \(\vec a\), \(\vec b\) and \(\vec c\) respectively. \(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and \(\vec {OC}\) = \(\vec c\).
M is the mid-point of AB.
Position vector of M (\(\vec {OM}\)) = \(\frac {\vec {OA} + \vec {OB}}2\) = \(\frac {\vec a + \vec b}2\)
Position vector of N(\(\vec {ON}\)) = \(\frac {\vec {OB} + \vec {OC}}2\) = \(\frac {\vec b + \vec c}2\)
Position vector of MN (\(\vec {MN}\)) = \(\vec {ON}\) - \(\vec {OM}\)
or, \(\vec {MN}\) = \(\frac {\vec b + \vec c}2\) - \(\frac {\vec a + \vec b}2\)
or, \(\vec {MN}\) = \(\frac {\vec b + \vec c - \vec a - \vec b}2\)
or, \(\vec {MN}\) = \(\frac {\vec c - \vec a}2\)
∴ 2\(\vec {MN}\) = \(\vec c\) - \(\vec a\)........................................(1)
\(\vec {AC}\) = \(\vec {OC}\) - \(\vec {OA}\) = \(\vec c\) - \(\vec a\).........................(2)
From (1) and (2)
\(\vec {AC}\) = 2\(\vec {MN}\) Hence, Proved
In the given figure, AB is the diameter of circle and O its centre. C is any point on the circumference of the circle. Prove by vector method that the \(\angle\)ACB = a right angle.
Let: O be the centre of a circle and AB be the diameter of the circle.
Let: C be any point in the semi-circle.
Let: \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OC}\) be the position vectors of A, B and C respectively with respect to the origin O.
Using triangle law of vector addition,
\(\vec {AC}\) = \(\vec {OC}\) - \(\vec {OA}\) and
\(\vec {BC}\) = \(\vec {OC}\) - \(\vec {OB}\)
∴ \(\vec {AC}\) . \(\vec {BC}\)
= (\(\vec {OC}\) - \(\vec {OA}\)) (\(\vec {OC}\) - \(\vec {OB}\))
= \(\vec {OC}\) . \(\vec {OC}\) - \(\vec {OC}\) . \(\vec {OB}\) - \(\vec {OA}\) . \(\vec {OC}\)+ \(\vec {OA}\) . \(\vec {OB}\) [\(\because\) \(\vec {OA}\) = - \(\vec {OB}\)]
= \(\vec {OC}^2\) + \(\vec {OC}\) . \(\vec {OA}\) - \(\vec {OC}\) . \(\vec {OA}\) - \(\vec {OB}^2\)
= \(\vec {OC}^2\) - \(\vec {OB}^2\)
= 0 [\(\because\) \(\begin {vmatrix} OC\\ \end {vmatrix}\) = \(\begin {vmatrix} OB\\ \end {vmatrix}\)]
∴ \(\vec {AC}\)⊥ \(\vec {BC}\)
Hence, \(\angle\)ACB = 90° Proved
Prove by vector method that the diagonals of a parallelogram bisect each other.
Let: ABCD be a parallelogram and M be the mid-point of the diagonal DB. Let: \(\vec {OA}\), \(\vec {OB}\), \(\vec {OC}\) and \(\vec {OD}\) be the position vectors of the points A, B, C and D of the parallelogram with reference to origin O.
Using mid-point theorem;
We have,
\(\vec {OM}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OD}\))
or, \(\vec {OM}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OA}\) + \(\vec {AD}\)) [\(\because\) \(\vec {OD}\) = \(\vec {OA}\) + \(\vec {AD}\)]
or, \(\vec {OM}\) = \(\frac 12\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {BC}\)) [\(\because\) \(\vec {AD}\) = \(\vec {BC}\)]
∴ \(\vec {OM}\) = \(\frac 12\) (\(\vec {OA}\) + \(\vec {OC}\) which is the position vector of the mid-point of \(\vec {AC}\)
∴ M is the mid-point of AC and BC.
∴ AC and BD bisect each other. Proved
If the position vector of the vertices A, B and C of a \(\triangle\)ABC are: \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OC}\) respectively, prove that the position vector of the centroid G of the triangle ABC is:
\(\vec g\) = \(\frac 13\)(\(\vec a\) + \(\vec b\) + \(\vec c\))
Let: \(\vec {OA}\) = \(\vec a\) , \(\vec {OB}\) = \(\vec b\), \(\vec {OC}\) = \(\vec c\) anf \(\vec {OG}\) = \(\vec g\)
The medians AE, BF and CD intersect at G. G is a centroid of the \(\triangle\)ABC.
Using mid-point formula,
In \(\triangle\)BOC,
\(\vec {OE}\) = \(\frac {\vec {OB} + \vec {OC}}2\).......................(1)
In any triangle, the centroid divides median in the ratio of 2 : 1.
In \(\triangle\)AOE,
\(\vec {OG}\) = \(\frac {1 . \vec {OA} + 2 . \vec {OE}}{1 + 2}\) [\(\because\) m : n = 2 : 1]
or, \(\vec g\) = \(\frac {\vec a + 2 . (\frac {\vec {OB} + \vec {OC}}2)}3\) [\(\because\) From eqn (1)]
or, \(\vec g\) = \(\frac {\vec a + \vec b + \vec c}3\)
∴\(\vec g\) = \(\frac13\)(\(\vec a\) + \(\vec b\) + \(\vec c\)) Proved
Prove vectorially that line joining vertex and mid-points of the base of an isosceles triangle is perpendicular to the base.
Let: PQR is an isosceles triangle where PQ = PR. M is the mid-ppointof QR.
Vertex P and M are joined.
To prove: PM⊥ QR
Using triangle law of vector addition:
\(\vec {QR}\) = \(\vec {QP}\) + \(\vec {PR}\)
\(\vec {QR}\) = - \(\vec {PQ}\) + \(\vec {PR}\) [\(\because\) \(\vec {QP}\) = - \(\vec {PQ}\)]
\(\vec {PM}\) = \(\frac 12\) (\(\vec {PQ}\) + \(\vec {PR})\) [\(\because\) M is the mid-point of QR]
Now,
\(\vec {PM}\) . \(\vec {QR}\)
= \(\frac 12\) (\(\vec {PQ}\) + \(\vec {PR}\)) (-\(\vec {PQ}\) + \(\vec {PR}\))
= \(\frac 12\) (\(\vec {PR}^2\) - \(\vec {PQ}^2\))
= \(\frac 12\) (PR2 - PQ2)
= \(\frac 12\) (PQ2 - PQ2) [\(\because\) PR = PQ]
= \(\frac 12\)× 0
= 0
Hence, PM⊥ QR.Proved
Prove by vector method that the perpendicular line is drawn from the vertex to the base of an isosceles triangle is bisect of the base.
Let: PQR is an isosceles triangle.
PM⊥ QR and PQ = PR
To prove: QM = MR
Using triangle law of vector addition,
In \(\triangle\)PQM,
\(\vec {PM}\) = \(\vec {PQ}\) + \(\vec {QM}\)........................(1)
In \(\triangle\)PRM,
\(\vec {PM}\) = \(\vec {PR}\) + \(\vec {RM}\)..........................(2)
From relation (1) and (2)
\(\vec {PQ}\) + \(\vec {QM}\) = \(\vec {PR}\) + \(\vec {RM}\)
or, \(\vec {PQ}\) + \(\vec {QM}\) = \(\vec {PQ}\) + \(\vec {RM}\) [\(\because\) \(\vec {PQ}\) = \(\vec {PR}\)]
∴\(\vec {QM}\) =\(\vec {RM}\)
Hence, PM bisect the line QR. Proved
Prove by vector method that medians of an isosceles triangle drawn from extremes of the base are equal.
Let: ABC is an isosceles triangle. CD and BE are medians. AB = AC
To prove: BE = CD
Let: \(\vec {BD}\) = \(\vec {DA}\) = \(\vec p\) and \(\vec {CE}\) = \(\vec {EA}\) = \(\vec q\)
Using triangle law of vector addition,
In \(\triangle\)BCE,
\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\)
\(\vec {BE}\) = \(\vec p\) + \(\vec p\) - \(\vec {EA}\) = 2\(\vec p\) - \(\vec q\).......................(1)
Squaring on both sides,
\(\vec {BE}^2\) = (2\(\vec p\) - \(\vec q\))2
\(\vec {BE}^2\) = 4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec q^2\)
\(\vec {BE}^2\) =4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\) [\(\because\) \(\vec p\) = \(\vec q\)]
\(\vec {BE}^2\) = 5\(\vec p^2\) - 4\(\vec p\)\(\vec q\)
In \(\triangle\)BCD,
\(\vec {CD}\) = \(\vec {CA}\) + \(\vec {AD}\) = \(\vec q\) + \(\vec q\) - \(\vec {DA}\) = 2\(\vec q\) - \(\vec p\)...................................(2)
Squaring on both sides,
\(\vec {CD}^2\) = (2\(\vec q\) - \(\vec p\))2
\(\vec {CD}^2\) = 4\(\vec q^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\)
\(\vec {CD}^2\) = 4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\) [\(\because\) \(\vec p\) = \(\vec q\)]
\(\vec {CD}^2\) = 5\(\vec p^2\) - 4\(\vec p\)\(\vec q\)
From above relation,
\(\vec {BE}^2\) = \(\vec {CD}^2\)
or, \(\vec {BE}\) = \(\vec {CD}\)
∴ BE = CD Proved
Prove by vector method that the triangle having two equal medians is an isosceles triangle.
Given: BE and CD are median of the \(\triangle\)ABC.
Let: \(\vec {BD}\) = \(\vec {DA}\) = \(\vec a\) , \(\vec {AE}\) = \(\vec {EC}\) = \(\vec b\)
To prove: AB = AC
In \(\triangle\)BAE,
\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\) = 2\(\vec a\) + \(\vec b\)
In \(\triangle\)ACD,
\(\vec {DC}\) = \(\vec {DA}\) + \(\vec {AC}\) = \(\vec a\) + 2\(\vec b\)
From question,
\(\vec {BE}\) = \(\vec {DC}\)
or, \(\vec {BE}^2\) = \(\vec {DC}^2\)
or, (2\(\vec a\) + \(\vec b\))2 = (\(\vec a\) + 2\(\vec b\))2
or, 4\(\vec a^2\) + 2\(\vec a\)\(\vec b\) + \(\vec b^2\) = \(\vec a^2\) + 4\(\vec a\)\(\vec b^2\) + 4\(\vec b^2\)
or, 4\(\vec a^2\) - \(\vec a^2\) + 4\(\vec a\)\(\vec b\) - 4\(\vec a\)\(\vec b\) = 4\(\vec b^2\) - \(\vec b^2\)
or, 3\(\vec a^2\) = 3\(\vec b^2\)
or, \(\begin {vmatrix} \vec a^2\\ \end {vmatrix}\) =\(\begin {vmatrix} \vec b^2\\ \end {vmatrix}\)
or,\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) =\(\begin {vmatrix} \vec b\\ \end {vmatrix}\)
or, 2\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = 2\(\begin {vmatrix} \vec b\\ \end {vmatrix}\)
∴AB = AC
Hence, ABC is an isosceles triangle. Proved
If a line is drawn from the centre of the circle to the mid-point of a chord prove by vector method that the line is perpendicular to the chord.
Let: O be the centre of the circle.
Join OA and OB.
Using triangle law of vector addition,
In \(\triangle\)AOC,
\(\vec {OA}\) = \(\vec {OC}\) + \(\vec {CA}\)
Squaring on both sides,
\(\vec {OA}^2\) = (\(\vec {OC}\) + \(\vec {CA}\))2 ................................(1)
In \(\triangle\)BOC,
\(\vec {OB}\) = \(\vec {OC}\) + \(\vec {CB}\)
\(\vec {OB}\) = \(\vec {OC}\) - \(\vec {CA}\) [\(\because\) \(\vec {CB}\) = -\(\vec {CA}\)]
Squaring on both sides,
\(\vec {OB}^2\) = (\(\vec {OC}\) - \(\vec {CA}\))2 ...............................(2)
Subtracting eqn (2) from (1)
\(\vec {OA}^2\) - \(\vec {OB}^2\) = (\(\vec {OC}\) + \(\vec {CA}\))2-(\(\vec {OC}\) - \(\vec {CA}\))2
or, \(\vec {OA}^2\) - \(\vec {OA}^2\) = \(\vec {OC}^2\) + \(\vec {CA}^2\) + 2\(\vec {OC}\) \(\vec {CA}\) - \(\vec {OC}^2\) - \(\vec {CA}^2\) + 2\(\vec {OC}\)\(\vec {CA}\)
or, 0 = 4\(\vec {OC}\)\(\vec {CA}\)
or, \(\vec {OC}\)\(\vec {CA}\) = 0
∴ \(\vec {OC}\)⊥ \(\vec {CA}\)
Hence, OC is perpendicular to AB. Proved
PQR is a triangle in which \(\angle\)Q = 90°, prove vectorially that:
\(\vec {PR}^2\) = \(\vec {PQ}^2\) + \(\vec {QR}^2\)
Let, PQR is a right angled triangle.
To prove:PR2 = PQ2 + QR2
Using triangle law of vector addition:
\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)
Squaring on both sides;
\(\vec {PR}^2\) = (\(\vec {PQ}\) + \(\vec {QR}\))2
or, \(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 . \(\vec {PQ}\) . \(\vec {QR}\) + \(\vec {QR}^2\)
or, \(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 × 0 + \(\vec {QR}^2\) [\(\because\) \(\vec {PQ}\) . \(\vec {QR}\) = 0]
∴ \(\vec {PR}^2\) = \(\vec {PQ}^2\) +\(\vec {QR}^2\) Proved
Derive \(\frac xa\) + \(\frac yb\) = 1 vectorially where the symbols have their usual meaning.
Let, AB be a straight line cuts x-axis at A(a, 0) and y-axis at B(0. b).
Let: P(x, y) be any point on the line AB.
\(\vec {OA}\) = \(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\)
\(\vec {OB}\) = \(\begin {pmatrix} 0\\ b\\ \end {pmatrix}\)
\(\vec {OP}\) = \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\)
\(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) =\(\begin {pmatrix} 0\\ b\\ \end {pmatrix}\) -\(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\) =\(\begin {pmatrix} -a\\ b\\ \end {pmatrix}\)
\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) -\(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\) =\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\)
\(\vec {AB}\) and \(\vec {AP}\) are parallel so,
\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\) = k\(\begin {pmatrix} -a\\ b\\ \end {pmatrix}\)
\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} -ka\\ kb\\ \end {pmatrix}\)
Taking corresponding elements of the equal vectors,
x - a = -ka
\(\frac xa\) = 1 - k .....................(1)
y = bk
\(\frac yb\) = k ............................(2)
Adding relation (1) and (2)
\(\frac xa\) +\(\frac yb\) = 1 - k + k
∴\(\frac xa\) +\(\frac yb\) = 1 Hence, Proved
Let, P be the point on the line determined by the point A and B such that AP : PB = m : n, then \(\vec {OP}\) = \(\frac {m.\vec {OB} + n.\vec {OA}}{m+n}\), the position vector of P in the relation to the origin O.
Proof:
Let, \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OP}\) be the position vectors of the points A, B and P relative to the origin O respectively.
From Question,
\(\frac {AP}{PB}\) = \(\frac mn\)
AP = \(\frac mn\)PB..........................(1)
Using vector of the above relation,
\(\vec {AP}\) = \(\frac mn\)\(\vec {PB}\)
\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) and
\(\vec {PB}\) = \(\vec {OB}\) - \(\vec {OP}\)
Putting the value of \(\vec {AP}\) and \(\vec {PB}\) in relation (1)
\(\vec {OP}\) - \(\vec {OA}\) = \(\frac mn\)(\(\vec {OB}\) - \(\vec {OP})\)
or, n \(\vec {OP}\) - n \(\vec {OA}\) = m \(\vec {OB}\) - m \(\vec {OP}\)
or,m \(\vec {OP}\) +n \(\vec {OP}\) = m \(\vec {OB}\) +n \(\vec {OA}\)
or, \(\vec {OP}\) (m + n) =m \(\vec {OB}\) +n \(\vec {OA}\)
∴ \(\vec {OP}\) = \(\frac {m.\vec {OB} + n.\vec {OA}}{m+n}\) Hence, Proved
Let, P be the external point on the line determined by the points A and B such that AP : PB = m : n such that: \(\vec {OP}\) = \(\frac {m. \vec {OB} - n. \vec {OA}}{m - n}\) the position vector of P in relation to the origin O.
Proof:
Let, \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OP}\) be the position vectors of the points A, B and P respectively in relation to the origin O.
From Question,
\(\frac {AP}{PB}\) = \(\frac mn\)
n. AP = m. PB
Using vector of the above relation,
n \(\vec {AP}\) = m \(\vec {PB}\)...........................(1)
From triangle law of vector addition:
\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) and
\(\vec {BP}\) = \(\vec {OP}\) - \(\vec {OB}\)
Putting the value of \(\vec {AP}\) and \(\vec {BP}\) in relation (1)
n (\(\vec {OP}\) - \(\vec {OA}\)) = m (\(\vec {OP}\) - \(\vec {OB}\))
or, n . \(\vec {OP}\) - n . \(\vec {OA}\) = m . \(\vec {OP}\) - m . \(\vec {OB}\)
or,n . \(\vec {OP}\) -m . \(\vec {OP}\) =n . \(\vec {OA}\)- m . \(\vec {OB}\)
or,\(\vec {OP}\) (n - m) =n . \(\vec {OA}\)- m . \(\vec {OB}\)
or,\(\vec {OP}\) = \(\frac {n . \vec {OA} - m . \vec {OB}}{n - m}\)
∴\(\vec {OP}\) = \(\frac {m. \vec {OB} - n. \vec {OA}}{m - n}\) Hence, Proved
OABC is a parallelogram, P and Q divide OC and CB, \(\vec {CP}\) : \(\vec {OP}\) = \(\vec {CQ}\) : \(\vec {QB}\) = 1 : 3. If \(\vec {OA}\) = \(\vec a\) and \(\vec {OC}\) = \(\vec c\), express PQ in terms of a and c and prove that: PQ and OB are parallel.
Given:
OABC is a parallelogram.
\(\vec {CP}\) : \(\vec {OP}\) = \(\vec {CQ}\) : \(\vec {QB}\) = 1 : 3
\(\vec {OA}\) = \(\vec a\) and \(\vec {OC}\) = \(\vec c\)
To prove: \(\vec {PQ}\)//\(\vec {OB}\) and find \(\vec {PQ}\) = ?
From Question:
\(\frac {\vec {CP}}{\vec {PO}}\) = \(\frac 13\) = \(\frac {\vec {CQ}}{\vec {QB}}\)
∴ \(\vec {PO}\) = 3 \(\vec {CP}\) and \(\vec {QB}\) = 3 \(\vec {CQ}\)
Using triangle law of vector addition:
\(\vec {OB}\) = \(\vec {OA}\) + \(\vec {AB}\)
or, \(\vec {OB}\)= \(\vec a\) + \(\vec c\)
or, \(\vec {OB}\) = \(\vec {OC}\) + \(\vec {CB}\)
or, \(\vec {OB}\) = \(\vec {OP}\) + \(\vec {PC}\) + \(\vec {CQ}\) + \(\vec {QB}\)
or, \(\vec {OB}\) = - \(\vec {PO}\) + \(\vec {PC}\) + \(\vec {CQ}\) + 3\(\vec {CQ}\)
or, \(\vec {OB}\) = -3 \(\vec {CP}\) - \(\vec {CP}\) + 4 \(\vec {CQ}\) [\(\because\) \(\vec {PO}\) = 3 \(\vec {CP}\)]
or, \(\vec {OB}\) = - 4 \(\vec {CP}\) + 4 \(\vec {CQ}\)
or, \(\vec {OB}\) = 4 \(\vec {PC}\) + 4 \(\vec {CQ}\)
or,\(\vec {OB}\) = 4 (\(\vec {PC}\) + \(\vec {CQ}\))
∴\(\vec {OB}\) = 4 \(\vec {PQ}\)
∴\(\vec {OB}\)//\(\vec {PQ}\) [\(\vec a\) = m\(\vec b\) then \(\vec a\)//\(\vec b\)]
and \(\vec {PQ}\) = \(\frac 14\)\(\vec {OB}\) = \(\frac 14\)(\(\vec a\) + \(\vec c\)) Proved
In the figure, ABCD is a parallelogram, L and M are mid-points of sides BC and CD respectively, prove that:
\(\vec {AL}\) + \(\vec {AM}\) = \(\frac 32\)\(\vec {AC}\)
Given:
ABCD is a parallelogram. L and M are mid-points of sides BC and CD respectively.
Using triangle law of vector addition:
\(\vec {AM}\) = \(\vec {AD}\) + \(\vec {DM}\) = \(\vec {AD}\) + \(\frac 12\)\(\vec {DC}\)...............(1)
\(\vec {AL}\) = \(\vec {AB}\) + \(\vec {BL}\) = \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\).....................(2)
Adding (1) and (2)
\(\vec {AM}\) +\(\vec {AL}\) =\(\vec {AD}\) + \(\frac 12\)\(\vec {DC}\) +\(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\)
or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {2 \vec {AD} + \vec {DC} + 2 \vec {AB} + \vec {BC}}2\)
or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {2 \vec {BC} + \vec {AB} + 2 \vec {AB} + \vec {BC}}2\) [\(\because\) \(\vec {AB}\) = \(\vec {DC}\) and \(\vec {AD}\) = \(\vec {BC}\)]
or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {3 \vec {AB} + 3 \vec {BC}}2\)
or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac 32\) (\(\vec {AB}\) + \(\vec {BC}\))
∴ \(\vec {AM}\) +\(\vec {AL}\) = \(\frac 32\)\(\vec {AC}\)
Hence, \(\vec {AL}\) +\(\vec {AM}\) = \(\frac 32\)\(\vec {AC}\) Proved
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