Note on Operations on Vector

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The operation by which two or more vectors or a scalar and a vector combine to give a single vector or a scalar is known as a vector operation. Here, in this level, we shall discuss the following operation,

  1. Addition of Vector
  2. Subtraction of vector
  3. Multiplication of a vector by a scalar.
Vector addition
Vector addition

Addition of vectors

Let \(\overrightarrow{a}\) = \(\begin{pmatrix}x_1\\y_1\\\end{pmatrix}\) and\(\overrightarrow{b}\) =\(\begin{pmatrix}x_2\\y_2\\\end{pmatrix}\), then \(\overrightarrow{a}\) +\(\overrightarrow{b}\) = \(\begin{pmatrix}x_1\\y_1\\\end{pmatrix}\) +\(\begin{pmatrix}x_2\\y_2\\\end{pmatrix}\) =\(\begin{pmatrix}x_1+x_2\\y_1+y_2\\\end{pmatrix}\).

\(\therefore\) x-component of \(\overrightarrow{a}\) +\(\overrightarrow{b}\) = x1 + x2

and, y-component of \(\overrightarrow{a}\) +\(\overrightarrow{b}\) = y1 + y2

Hence, the addition of two column vectors is obtained by adding the corresponding x components and y components of \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

Triangle Law of Vector Addition

Let \(\overrightarrow{AB}\) displace point A to point B \(\overrightarrow{BC}\) displace point B to point C. Then, the total displacement from A to C i.e. \(\overrightarrow{AC}\) is given by\(\overrightarrow{AC}\) +\(\overrightarrow{BC}\).

This law of addition of vector is known as triangle law of vector addition and\(\overrightarrow{AC}\) is called the resultant vector.

Hence, triangle law of vector addition states that if two sides of a triangle taken in order tp represent two vectors then theirresultant will be given by the third side of the triangle in opposite order.

Parallelogram Law of a vector
1
Parallelogram Law of a vector
Source:mathinsight.org

Let \(\overrightarrow{AB}\) = \(\overrightarrow{a}\) and \(\overrightarrow{AD}\) = \(\overrightarrow{b}\) be two co-initial (having the same point of start) vectors with the initial points as A. A parallelogram ABCD is completed such that AC = DC and AD = BC. Also, we have , AB//CD and AD//BC.

\(\therefore\) \(\overrightarrow{DC}\) = \(\overrightarrow{AB}\) = \(\overrightarrow{a}\) and

\(\overrightarrow{AD}\) = \(\overrightarrow{BC}\) = \(\overrightarrow{b}\)

Now, by triangle law of vector addition we have,

\(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) = \(\overrightarrow{AC}\)

or, \(\overrightarrow{AB}\) + \(\overrightarrow{AD}\) = \(\overrightarrow{AC}\)(\(\therefore\) \(\overrightarrow{AD}\) = \(\overrightarrow{BC}\))

\(\therefore\) \(\overrightarrow{a}\) + \(\overrightarrow{b}\) = \(\overrightarrow{AC}\)

This law of addition of vector is called parallelogram law of vector addition. Hence, the parallelogram law of vector adition states that if two adjacent sides of a parallelogram represent two co-initial vectors then the diagonal of the parallelogram passing through the same point gives its resultant.

Polygon Law of Vector Addition
Polygon Law of Vector Addition
Source:www.tutorvista.com
Polygon Law of Vector Addition

Polygon law of vector addition is the generalization of the triangle law of vector addition where \(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) + \(\overrightarrow{CD}\) + \(\overrightarrow{DE}\) = \(\overrightarrow{AE}\).

Proof:

ABCDE is a polygon (pentagon). Let's join AC and AD.

Now using \(\triangle\) law of vector addition in \(\triangle\)ABC, \(\triangle\)ADC and \(\triangle\)ADE, we get

\(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) = \(\overrightarrow{AC}\).......(1)

\(\overrightarrow{AC}\) + \(\overrightarrow{CD}\) = \(\overrightarrow{AD}\).........(2)

Using (1) and (2)

\(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) + \(\overrightarrow{CD}\) = \(\overrightarrow{AD}\).......(3)

Again, \(\overrightarrow{AD}\) + \(\overrightarrow{DE}\) = \(\overrightarrow{AE}\) .....()

using (3) and (4)

\(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) + \(\overrightarrow{CD}\) + \(\overrightarrow{AD}\) + \(\overrightarrow{DE}\) = \(\overrightarrow{AE}\) proved.

Hence, the polygon law of vector addition states that if a number of vectors are represented by the sides of a polygon taken in order then the resultant vector is represented by the closing side of the polygon in the opposite order.

Subtraction of vectors

Subtraction of vectors
Source:tutorial.math.lamar.edu
Subtraction of vectors

Let \(\overrightarrow{a}\) = \(\begin{pmatrix}a_1\\a_2\\\end{pmatrix}\) and\(\overrightarrow{b}\) =\(\begin{pmatrix}b_1\\b_2\\\end{pmatrix}\), then \(\overrightarrow{a}\) - \(\overrightarrow{b}\) = \(\begin{pmatrix}a_1\\b_2\\\end{pmatrix}\) - \(\begin{pmatrix}b_1\\b_2\\\end{pmatrix}\) =\(\begin{pmatrix}a_1-a_2\\b_1-b_2\\\end{pmatrix}\).

Subtraction is a reverse of addition. So. if AB and BC are two vectors then their difference of \(\overrightarrow{AB}\) - \(\overrightarrow{BC}\) is the sum of the vectors \(\overrightarrow{AB}\) and the negative of\(\overrightarrow{BC}\) i.e. -\(\overrightarrow{BC}\)

So,

\(\overrightarrow{AB}\) - \(\overrightarrow{BC}\) = \(\overrightarrow{AB}\) + \(\overrightarrow{BC}\)

Here,

\(\overrightarrow{AB}\) + \(\overrightarrow{BC}\) =\(\overrightarrow{AC}\)

But \(\overrightarrow{BD}\) is negative of \(\overrightarrow{BC}\).

i.e \(\overrightarrow{BD}\) = -\(\overrightarrow{BC}\)

or, \(\overrightarrow{AB}\) - \(\overrightarrow{BC}\)

= \(\overrightarrow{AB}\) + (-\(\overrightarrow{BC}\))

= \(\overrightarrow{AB}\) + \(\overrightarrow{BD}\)

= \(\overrightarrow{AD}\)

Multiplication of a vector by a scalar

Multiplication of a vector by a scalar

\(\overrightarrow{a}\) = \(\begin{pmatrix}a_1\\a_2\\\end{pmatrix}\) k \(\overrightarrow{a}\) = k\(\begin{pmatrix}a_1\\a_2\\\end{pmatrix}\) =\(\begin{pmatrix}ka_1\\ka_2\\\end{pmatrix}\) where k is a scalar.

If k is a positive scalar, \(\overrightarrow{a}\) and \(\overrightarrow{a}\) have same direction. If k is a negative scalar\(\overrightarrow{a}\) and k\(\overrightarrow{a}\) will have opposite direction.

  1. The addition of two column vectors is obtained by adding the corresponding x components and y components of \(\overrightarrow{a}\) and \(\overrightarrow{b}\).
  2. Subtraction is a reverse of addition.
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Very Short Questions

Here , given , \(\overrightarrow {AB}\) = \(\begin {pmatrix} -2 \\ -5 \end {pmatrix}\) , \(\overrightarrow {CD}\) = \(\begin {pmatrix} -6 \\ -15 \end{pmatrix}\)
Here , \(\overrightarrow {CD}\) = \(\begin {pmatrix} -6 \\ -15 \end {pmatrix}\) = \(\begin {pmatrix} 3× -2 \\ 3 × -5 \end{pmatrix}\) = 3 \(\begin {pmatrix} -2 \\ -5 \end{pmatrix}\) = 3\(\overrightarrow {AB}\)
So , \(\overrightarrow {CD}\) = 3\(\overrightarrow {AB}\)
\(\therefore\) \(\overrightarrow {CD}\) and \(\overrightarrow {AB}\) are parallel. Proved.

Here , given \(\overrightarrow {IJ}\) = \(\begin {pmatrix} 15 \\ -12 \end{pmatrix}\) , \(\overrightarrow {KL}\) = \(\begin {pmatrix} 5 \\ -4 \end{pmatrix}\)
Here , \(\overrightarrow {IJ}\) = \(\begin {pmatrix} 15 \\ -12 \end{pmatrix}\) = \(\begin {pmatrix} 3× 5 \\ 3× -4 \end {pmatrix}\) = 3 \(\begin {pmatrix} 5 \\ -4 \end {pmatrix}\) = 3 \(\overrightarrow {KL}\)
\(\therefore\) \(\overrightarrow {IJ}\) and \(\overrightarrow {KL}\) are parallel. Proved.

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  • (overrightarrow{a}) (egin{pmatrix}4\ 5 end{pmatrix})  and (overrightarrow{b})  (egin{pmatrix}10\ -3 end{pmatrix}) find /(overrightarrow{a})+2(overrightarrow{b})/.

    40.01 units


    24.02 units


    26.14 units


    54.06 units


  • (overrightarrow{a}) (egin{pmatrix}4\ 5 end{pmatrix})  and (overrightarrow{b})  (egin{pmatrix}10\ -3 end{pmatrix}) find /(overrightarrow{a})+2(overrightarrow{b})/.

    54.06 units


    26.14 units


    24.02 units


    40.01 units


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    34.17 units


    14.87 units


    47.22 units


    14.87 units


  • (overrightarrow{a}) (egin{pmatrix}6\ 2 end{pmatrix}),  and (overrightarrow{b})  (egin{pmatrix}2\ 3end{pmatrix}) and (overrightarrow{c})= (egin{pmatrix}10\ 8 end{pmatrix}) (overrightarrow{a})+(overrightarrow{xb})=(overrightarrow{c}),find  the value of the scalar x.

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  • (overrightarrow{a}) (egin{pmatrix}2\ 1 end{pmatrix}),  and (overrightarrow{b})  (egin{pmatrix}-1\ 2end{pmatrix}) and (overrightarrow{c})= (egin{pmatrix}-1\ 7 end{pmatrix}), find λ such that:(overrightarrow{a})+λ(overrightarrow{b})=(overrightarrow{c})

    3


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DISCUSSIONS ABOUT THIS NOTE

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RK COMPUTER

PROVE THAT- VECTOR AB AC AD EA FA=4AB


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Lattu Mahato

ABCDEF is a regular poligon.Vector AB vectorAC vector AD vector EA vector FA = vector 4AB


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