Vectors in math are objects that have both magnitude and direction. To represent math vectors, we use an arrow having a line of given length and pointing along a given direction. The magnitude of a vector is denoted by its length. |

Some examples are length, area, volume, temperature, mass and time. In vectors, a ordinary number is called a scalar.

**Example:**7, - 0.5656, 89, -96 etc. → Read More A unit vector is a vector having magnitude and direction. Magnitude of a unit vector is 1. Any vector can be expressed in terms of unit vector.

A unit vector is represented as $\hat{U}$ = $\frac{\vec{U}}{|\vec{U}|}$

where, U is the given vector and |U| is the magnitude of the given vector.

Magnitude of a vector is the length of the vector. It cannot be added algebraically. If A is the vector, then the length of $\vec{A}$ is denoted by |A|.

|A| = $\sqrt{l^{2} + m^{2}}$ is the formula for finding the magnitude of the vector, where, l and m are the components of the vector $\vec{A}$.

Two vectors a and b are parallel, if their cross product is zero. If the vectors have the same direction, then vectors are said to be parallel.

Components of one vector must be in the same ratio, when compared to the corresponding components of the parallel vector.

Zero vector is a vector, which has zero magnitude and an arbitrary direction. It is a vector of length 0 and thus, has all components equal to zero. Zero vector is represented as $\vec{0}$. In a zero vector, the initial point coincides with the terminal point and hence, its direction is indeterminate.

When a Zero vector is added to another vector $\vec{a}$, the result is $\vec{a}$ only.

$\vec{a}$ + $\vec{0}$ = $\vec{a}$

**Examples:**The position vector of the origin of the coordinate axes is a zero vector. The displacement of a ball thrown up and received back by the thrower is a zero vector.

Cross product of two vectors is the area of the parallelogram between them and a cross product has both magnitude and direction.

Let $\vec{a}$ and $\vec{b}$ be any two vectors. Thus, their cross product denoted by $\vec{a}$ x $\vec{b}$ is defined by $\vec{a}$ x $\vec{b}$ = |$\vec{a}$| |$\vec{b}$| sin $\theta$ $\hat{n}$

where, $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ such that, $\vec{a}$, $\vec{b}$ and $\hat{n}$ form a right handed system.

$\vec{a}$ x $\vec{b}$ is a vector and hence, the name vector product or cross product.

The name cross product is due to the notation of placing a cross between $\vec{a}$ and $\vec{b}$ while denoting the cross product.

|$\vec{a}$ x $\vec{b}$| = |$\vec{a}$| |$\vec{b}$| sin $\theta$; |$\hat{n}$| = 1 If $\vec{a}$ and $\vec{b}$ are two vectors, then their scalar product or dot product denoted by $\vec{a}$ . $\vec{b}$ is defined by

$\vec{a}$ . $\vec{b}$ = |$\vec{a}$| . |$\vec{b}$| cos $\theta$

where, $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

The name scalar product is due to the fact that $\vec{a}$ . $\vec{b}$ is a scalar and the name dot product is due to the notation of placing a dot between the two vectors while denoting the scalar product.

Suppose a man is pulling a body with a force ‘a’ and the body is rolling horizontally on the ground, and we are asked that if the man takes that body to another point say, ‘b’ then how much work is done?

We know that, the work done on a body is the multiplication of force applied to the body and distance traveled. In the above case, the work done will be the scalar product of a and b. So, the work done will be,

$\vec{a}$ . $\vec{b}$ = |$\vec{a}$|.|$\vec{b}$| cos $\theta$

If the value of angle $\theta$ is 90$^o$, then the work done will be zero as cos 90$^o$ = 0. If the angle is zero degree, then the work is maximum because cos 0$^o$ = 1. So, as the angle will decrease, the work done will be increased.

Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be any three vectors. Then, $\vec{a}$ x ($\vec{b}$ x $\vec{c}$) is called a vector triple product and this is the vector product of $\vec{a}$ with the vector $\vec{b}$ x $\vec{c}$ and hence, the result is a vector. We can define vector triple product such as $\vec{b}$ x ($\vec{a}$ x $\vec{c}$), $\vec{c}$ x ($\vec{a}$ x $\vec{b}$) etc.

Let us consider the vector triple product $\vec{a}$ x ($\vec{b}$ x $\vec{c}$). This being the vector product of $\vec{a}$ and ($\vec{b}$ x $\vec{c}$), is perpendicular to both the vectors $\vec{a}$ x ($\vec{b}$ x $\vec{c}$). But, ($\vec{b}$ x $\vec{c}$) is perpendicular to both $\vec{b}$ and $\vec{c}$. Hence, the three vectors $\vec{a}$ x ($\vec{b}$ x $\vec{c}$), $\vec{b}$ and $\vec{c}$ are perpendicular to the same vector ($\vec{b}$ x $\vec{c}$). Therefore, $\vec{a}$ x ($\vec{b}$ x $\vec{c}$) is coplanar with $\vec{b}$ and $\vec{c}$. Hence, $\vec{a}$ x ($\vec{b}$ x $\vec{c}$) is a vector in the plane determined by $\vec{b}$ and $\vec{c}$ and is also perpendicular to $\vec{a}$. It is to be noted that $\vec{a}$ x ($\vec{b}$ x $\vec{c}$) $\neq$ ($\vec{a}$ x $\vec{b}$) x $\vec{c}$. Therefore, in writing the vector triple product, parenthesis are extremely important. Vectors are added according to the triangle law of addition. We proceed as follows:

Let $\vec{a}$ and $\vec{b}$ be two vectors to be added. Choose an arbitrary point O. Take O as the initial point of $\vec{a}$ and A be its terminal point so that, $\vec{OA}$ = $\vec{a}$. Place the initial point of $\vec{b}$ on the terminal point A of $\vec{a}$. Let the terminal point of $\vec{b}$ now be B, so that $\vec{AB}$ = $\vec{b}$. Draw vector $\vec{OB}$, which is sum of $\vec{a}$ and $\vec{b}$ and we write $\vec{OB}$ = $\vec{a}$ + $\vec{b}$

Complete the parallelogram OABC. Clearly, OA = CB and $\vec{OA}$ and $\vec{CB}$ are equal vectors. Similarly, $\vec{OC}$ and $\vec{AB}$ are equal vectors.

Therefore, $\vec{OA}$ = $\vec{a}$, $\vec{OC}$ = $\vec{AB}$ = $\vec{b}$. Now, from the figure, we see that $\vec{a}$ + $\vec{b}$ = $\vec{OB}$ is the vector along the diagonal of the parallelogram, whose adjacent sides are $\vec{a}$ and $\vec{b}$. Hence, the triangle law of addition is sometimes called the parallelogram law of vector addition. That is, if two vectors $\vec{a}$ and $\vec{b}$ acting at a point are represented by the two adjacent sides of a parallelogram, then their resultant is given by the diagonal of the parallelogram passing through that point.

The following laws are satisfied by vector addition:

$\vec{a}$ + $\vec{b}$ = $\vec{b}$ + $\vec{a}$ (Commutative law)

If $\vec{a}$, $\vec{b}$, $\vec{c}$ are any three vectors, then $\vec{a}$ + ($\vec{b}$ + $\vec{c}$) = ($\vec{a}$ + $\vec{b}$) + $\vec{c}$ (Associative law)

$\vec{a}$ + $\vec{0}$ = $\vec{0}$ + $\vec{a}$ = $\vec{a}$ (Additive Identity law) If $\vec{a}$ and $\vec{b}$ are any two vectors, then $\vec{a}$ - $\vec{b}$ is defined as $\vec{a}$ + (- $\vec{b}$). $\vec{a}$ - $\vec{b}$ is the result of subtraction of $\vec{b}$ from $\vec{a}$. Thus, $\vec{a}$ - $\vec{b}$ = $\vec{a}$ + (-$\vec{b}$). This operation is called as subtraction of two vectors.

In the subtracting operation, we consider only the magnitude of the vector not the direction.

Let $\vec{OA}$ = $\vec{a}$, -$\vec{b}$ = $\vec{AB}$

Complete the parallelogram OABC.

$\vec{OA}$ + $\vec{AB}$ = $\vec{OB}$

-$\vec{b}$ + $\vec{a}$ = $\vec{a}$ - $\vec{b}$ = $\vec{OB}$

Then, in the parallelogram OABC with $\vec{a}$ and -$\vec{b}$ as adjacent sides, $\vec{OB}$ = $\vec{a}$ - $\vec{b}$ Vector multiplication is one of the several techniques for the multiplication of two vectors with themselves. It can be dot product, cross product, scalar vector multiplication etc. Given below are some of the examples of vector multiplication.

### Solved Examples

**Question 1:**Find the cross product between a = (2, -4, 6) and b = (2, 7, 9).

**Solution:**

**Given:**a = (2, -4, 6) and b = (2, 7, 9)

Now, we need to find the cross product of a and b

Cross product is as follows:

a x b = $\begin{bmatrix}

i & j & k\\

2&-4 & 6\\

2& 7& 9

\end{bmatrix}$

= i(-4.9 - 7.6) -j(2.9 - 2.6) + k(2.7 + 2.4)

= i( -36 - 42) -j(18 - 12) + k(14 + 8)

= -78i -6j + 22k

**Question 2:**Find the angle between a = 5i + 2j + 3k and b = 3i + 2j + k

**Solution:**

|a| = $\sqrt{25 + 4 + 9}$ = $\sqrt{38}$

|b| = $\sqrt{9 + 4 + 1}$ = $\sqrt{14}$

a . b = 5 . 3 + 2 . 2 + 3 . 1

= 15 + 4 + 3 = 22

$\theta$ = cos$^{-1}$ [$\frac{22}{\sqrt{38}\sqrt{14}}$]