## 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,

2. Subtraction of vector
3. Multiplication of a vector by a scalar.

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

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 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

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

$$\overrightarrow{a}$$ = $$\begin{pmatrix}a_1\\a_2\\\end{pmatrix}$$ 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.
.

### 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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