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

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

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) = 8.11⁰_Ans

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) = 29.54⁰_Ans

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 \(\vec {AC}\).

\(\vec {AB}\) + \(\vec {AD}\) = \(\vec {AC}\)

\(\vec u\) + \(\vec v\) = \(\vec {AC}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\vec {AB}\), \(\vec {BC}\) 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 \(\triangle\)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

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

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

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

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

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

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 + RS²_{Hence, Proved}

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

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

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}

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}

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}

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}

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}

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

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

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

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}

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}

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

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

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

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

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

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

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

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

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

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}

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}

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}

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

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