## What is Vector Arithmetic?

Vectors are those objects which have a magnitude along with the direction. In vector arithmetic, we will see how arithmetic operators like addition and multiplication are used on any two vectors. Arithmetic in basic means dealing with numbers. Here, magnitude means the length or the size of an object. The notation used is the arrow over the head of the vector indicating its direction.

Consider two vectors a and b. When will you say these two are equal? They are equal if both of them have the same length and direction. Here, the length is termed as the magnitude.

## Operations

Suppose that you and two of your friends named Jim and Jack are playing throw ball. Three of you are standing in a right triangle. On scaling, the distance between you and Jim was found to be 9 meters and the distance between you and Jack is 8 meters. What is the distance between Jim and Jack?

Now you have to see this as vectors. One has magnitude 9 and it is in the direction towards the vector named Jim. The other has magnitude 8 and it is in the direction towards the vector named Jack. Since this forms a right-angled triangle, we can use the Pythagoras theorem. Thus, the distance between Jack and Jim is the square root (sqrt) of the sum of the squares of the two vectors. The new vector is the distance which is 12.04 meters. In this way, addition is done.

Addition can be in any order. If ‘a’ and ‘b’ are any two vectors, adding ‘a’ and ‘b’ or ‘b’ and ‘a’ will give the same value.

### Subtraction

Suppose that your friend is walking towards you. This means that your friend is coming from the opposite direction. This is indicated by a negative sign. Can you tell the difference between vector ‘a’ and ‘-a’. They both have the same magnitude. But the direction is the opposite. When you add these two, you get the new vector called the zero vector. Adding a negative and a positive term is equal to subtracting two terms. In this way, subtraction is done.

### Multiply by Scalars

Scalars are those quantities that have only magnitude. They can be any real numbers. What happens when you multiply a vector by a scalar?

To multiply a vector by a scalar, multiply each component of the vector by that scalar.

Suppose that the scalar is any value greater than 1. If a vector is multiplied by this scalar, the magnitude of the vector increases.

For example, let the vector be a and the scalar be any value greater than 1. On multiplication, the magnitude increases for any positive scalar. Also, it does not affect the direction of the vector.

Suppose that the scalar is any value between 0 and 1. If a vector is multiplied by this scalar, the magnitude of the vector decreases but does not affect the direction of the vector.

If the scalar is of any value less than 0, the direction of the vector gets changed.

### Properties of Scalar Multiplication

Let the letters ‘s’ and ‘t’ denote two scalars and ‘a’ and ‘b’ denote two vectors.

• If ‘s’ is multiplied by the sum of ‘a’ and ‘b’, the result is the sum of ‘s’ multiplied by ‘a’ and ‘s’ multiplied by ‘b’. It is mathematically represented as $s\left(a+b\right)=sa+sb$
• If ‘a’ is multiplied by the sum of ‘s’ and ‘t’, the result is the sum of ‘a’ multiplied by ‘s’ and ‘a’ multiplied by ‘t’. It is mathematically represented as $\left(s+t\right)a=sa+ta$
• Let s = 1. Then, multiplying ‘s’ by ‘a’ gives ‘a’. It is mathematically represented as $1\left(a\right)=a$
• Let s = 0. Then, multiplying ‘0’ by ‘a’ gives ‘0’. It is mathematically represented as $0\left(a\right)=0$
• Let s = -1. Then, multiplying ‘s’ by ‘a’ gives ‘-a’. It is mathematically represented as $\left(-1\right)a=-a$

You can prove if the given properties are true by calculating the left side and the right side of the equation. If the left side and right side are equal then it is true.

## Problems

1. Let s = 3, a = (2,1) and b = (5,7). Show that $s\left(a+b\right)=sa+sb$

Let us first calculate the left side of the equation.

$\begin{array}{c}s\left(a+b\right)=3\left(\left(2,1\right)+\left(5,7\right)\right)\\ =3\left(7,8\right)\\ =\left(21,24\right)\end{array}$

Now, calculate the right side of the equation.

$\begin{array}{c}sa+sb=3\left(2,1\right)+3\left(5,7\right)\\ =\left(6,3\right)+\left(15,21\right)\\ =\left(21,24\right)\end{array}$

Here, the left side and right side are equal.

Thus, $s\left(a+b\right)=sa+sb$

2. Let s = 4, a = (1,1) and b = (3,5). Can you find if $s\left(a+b\right)=sa+sb$?

Let us first calculate the left side of the equation.

$\begin{array}{c}s\left(a+b\right)=4\left(\left(1,1\right)+\left(3,5\right)\right)\\ =4\left(4,6\right)\\ =\left(16,24\right)\end{array}$

Now, calculate the right side of the equation.

$\begin{array}{c}sa+sb=4\left(1,1\right)+4\left(3,5\right)\\ =\left(4,4\right)+\left(12,20\right)\\ =\left(16,24\right)\end{array}$

Here, the left side and right side are equal.

Thus, $s\left(a+b\right)=sa+sb$.

3. Let s = 3, t = 4 and a = (5,6). Show that $\left(s+t\right)a=sa+ta$.

Let us first calculate the left side of the equation.

$\begin{array}{c}\left(s+t\right)a=\left(3+4\right)\left(5,6\right)\\ =7\left(5,6\right)\\ =\left(35,42\right)\end{array}$

Now, calculate the right side of the equation.

$\begin{array}{c}sa+ta=3\left(5,6\right)+4\left(5,6\right)\\ =\left(15,18\right)+\left(20,24\right)\\ =\left(35,42\right)\end{array}$

Here, the left side and right side are equal.

Thus, $\left(s+t\right)a=sa+ta$

4. Let s = 4, t = 6 and a = (2,8) . Can you find if $a\left(s+t\right)=as+at$?

Let us first calculate the left side of the equation.

$\begin{array}{c}a\left(s+t\right)=\left(2,8\right)\left(4+6\right)\\ =\left(2,8\right)10\\ =\left(20,80\right)\end{array}$

Now, calculate the right side of the equation.

$\begin{array}{c}as+at=\left(2,8\right)4+\left(2,8\right)6\\ =\left(8,32\right)+\left(12,48\right)\\ =\left(20,80\right)\end{array}$

Here, the left side and right side are equal.

Thus, $a\left(s+t\right)=as+at$.

## Multiplication

There are two ways in which vectors can be multiplied. One is the cross product and the other is the dot product.

### Cross Product

It is represented by a cross sign between two vectors. Taking the cross product of the two gives a new vector.

It is given by the formula $a×b=\left|a\right|\left|b\right|sin\theta \stackrel{^}{n}$.

Here, $\left|a\right|$ is the magnitude of vector a, $\left|b\right|$ is the magnitude of vector b and $\theta$ is the angle between a and b. Also, $\stackrel{^}{n}$ is a unit vector perpendicular to both a and b. It shows the direction.

A unit vector is a vector with a magnitude equal to 1.

If two vectors are written in component form as $a=\left({a}_{1},{a}_{2},{a}_{3}\right)$ and $b=\left({b}_{1},{b}_{2},{b}_{3}\right)$, then the cross product is defined as $a×b=\left|\begin{array}{ccc}i& j& k\\ {a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\end{array}\right|$, where i, j, and k are the standard unit vectors in three dimensions.

Example:

1. Find the cross product between $a=\left(3,-3,1\right)$ and $b=\left(4,1,2\right)$

The cross product is given by the following matrix.

$a×b=\left|\begin{array}{ccc}i& j& k\\ 3& -3& 1\\ 4& 1& 2\end{array}\right|$

Let i, j, and k be the standard unit vectors in three dimensions.

We must find the determinant of the matrix.

$\begin{array}{c}\left|\begin{array}{ccc}i& j& k\\ 3& -3& 1\\ 4& 1& 2\end{array}\right|=i\left(\left(-3\cdot 2\right)-\left(1\cdot 1\right)\right)-j\left(\left(3\cdot 2\right)-\left(1\cdot 4\right)\right)+k\left(\left(3\cdot 1\right)-\left(-3\cdot 4\right)\right)\\ =i\left(-6-1\right)-j\left(6-4\right)+k\left(3+12\right)\\ =-7i-2j+15k\end{array}$

The result is a vector quantity.

2. Can you find the cross product between $a=\left(1,-3,1\right)$ and $b=\left(4,9,2\right)$?

### Dot Product

It is represented by a dot sign between two vectors. Taking the dot product of the two gives a scalar.

It is given by the formula $a.b=\left|a\right|\left|b\right|\mathrm{cos}\theta$, where, $\left|a\right|$ is the magnitude of vector a, $\left|b\right|$ is the magnitude of vector b and $\theta$ is the angle between a and b.

The component formula for the dot product is given as follows:

$\left({a}_{1},{a}_{2},{a}_{3}\right)\cdot \left({b}_{1},{b}_{2},{b}_{3}\right)={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}$.

Let us see an example.

1. Find the dot product of $a=\left(1,2,3\right)$ and $b=\left(4,-5,6\right)$

We will use the component formula $\left({a}_{1},{a}_{2},{a}_{3}\right)\cdot \left({b}_{1},{b}_{2},{b}_{3}\right)={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}$.

Compute the dot product using the above formula. The calculation is given below.

$\begin{array}{c}1\left(4\right)+2\left(-5\right)+3\left(6\right)=4-10+18\\ =12\end{array}$

The result is a scalar quantity.

2. Can you find the dot product of $a=\left(1,-3,1\right)$ and $b=\left(4,9,2\right)$?

Use the component formula $\left({a}_{1},{a}_{2},{a}_{3}\right)\cdot \left({b}_{1},{b}_{2},{b}_{3}\right)={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}$ to calculate the dot product.

$\begin{array}{c}\left(1,-3,1\right)\cdot \left(4,9,2\right)=1\left(4\right)+\left(-3\right)\left(9\right)+1\left(2\right)\\ =4-27+2\\ =-21\end{array}$

The result is a scalar quantity.

## Parallel

Two vectors are said to be parallel if one is the scalar multiple of the other.

## Perpendicular

Two vectors are said to be perpendicular if their dot product is equal to zero.

## Formula

• The formulas for the cross product of two vectors are:

$a×b=\left|a\right|\left|b\right|sin\theta \stackrel{^}{n}$

$a×b=\left|\begin{array}{ccc}i& j& k\\ {a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\end{array}\right|$

• The formulas for the dot product of two vectors are:

$a.b=\left|a\right|\left|b\right|\mathrm{cos}\theta$

$a\cdot b={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}$

Here, $\left|a\right|$ is the magnitude of vector a, $\left|b\right|$ is the magnitude of vector b, $\theta$ is the angle between a and b, and $\stackrel{^}{n}$ is a unit vector perpendicular to both a and b. Also, $a=\left({a}_{1},{a}_{2},{a}_{3}\right)$ and $b=\left({b}_{1},{b}_{2},{b}_{3}\right)$ are the component forms of the vectors a and b.

## Practice Problems

Use the given operation to add or subtract the given vectors.

1. Add $a=\left(1,3,1\right)$ and $b=\left(4,9,2\right)$.

Use the formula $\left({a}_{1},{b}_{1},{c}_{1}\right)+\left({a}_{2},{b}_{2},{c}_{2}\right)=\left({a}_{1}+{a}_{2},{b}_{1}+{b}_{2},{c}_{1}+{c}_{2}\right)$ to add the vectors.

$\left(1,3,1\right)+\left(4,9,2\right)=\left(5,12,3\right)$

(Hint: You have to do it component wise)

2. Subtract $b=\left(5,3,2\right)$ from $a=\left(9,3,7\right)$.

Use the formula $\left({a}_{1},{b}_{1},{c}_{1}\right)-\left({a}_{2},{b}_{2},{c}_{2}\right)=\left({a}_{1}-{a}_{2},{b}_{1}-{b}_{2},{c}_{1}-{c}_{2}\right)$ to add the vectors.

$\left(9,3,7\right)-\left(5,3,2\right)=\left(4,0,5\right)$

## Context and Applications

Vector Arithmetic is used in the K-12 curriculum, undergraduate and post-graduate mathematics, and physics and is asked in entrance examinations.

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

• Bachelors in Mathematics
• Masters in Mathematics

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