In linear algebra, a real number is considered as scalar, and we can also relate the term scalar with the vectors in vector space by using the operation known as Scalar Multiplication. In this operation a scalar or a number is multiplied by a vector to get back another vector. |

Thus if $\vec{a}$ is a vector, then -3 $\vec{a}$ is a vector whose magnitude is |-3| |$\vec{a}$| = 3 |$\vec{a}$| and whose direction is opposite to that of $\vec{a}$.

Scalar multiplication satisfies the following properties:

If m and n are any scalars and $\vec{a}$ and $\vec{b}$ are vectors then

- (m + n ) $\vec{a}$ = m$\vec{a}$ + n$\vec{a}$
- m($\vec{a}$ + $\vec{b}$) = m$\vec{a}$ + m$\vec{b}$
- m(n$\vec{a}$ ) = n(m$\vec{a}$ ) = mn $\vec{a}$

Vector cross product is defined as

A * B = AB sin$\theta$ $\hat{n}$The vector $\hat{n}$ is a unit vector perpendicular to the plane formed by the two vectors. Cross product is distributive.

Cross product of any vector with itself is zero. That is, A * A = 0 Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three given vectors. Then $\vec{a}$ . ($\vec{b}$ * $\vec{c}$) is called a scalar triple product. It is the scalar product of the vectors a and ($\vec{b}$ * $\vec{c}$) and hence the result is a scalar. With $\vec{a}$, $\vec{b}$, $\vec{c}$ we can also from scalar triple product such as $\vec{b}$ . ($\vec{c}$ * $\vec{a}$), $\vec{c}$ . ($\vec{a}$ * $\vec{b}$)

Scalar triple product can be evaluated in two steps by first calculating b * c followed by a dot product with a or as below.

a. b * c = $\begin{vmatrix}

a_{1} & a_{2} &a_{3} \\

b_{1} &b_{2} &b_{3} \\

c_{1}&c_{2} &c_{3}

\end{vmatrix}$

Scalar triple product has the following properties.

- $\vec{a}$ . ( $\vec{b}$ * $\vec{c}$) = $\vec{b}$ . $\vec{c}$ * $\vec{a}$ = $\vec{c}$ . ($\vec{a}$ * $\vec{b}$)
- Scalar product is zero if any two vectors are identical.
- If $\vec{a}$, $\vec{b}$, $\vec{c}$ are coplanar then $\vec{a}$ . $\vec{b}$ * $\vec{c}$ = 0 and conversely.

### Solved Example

**Question:**Find the scalar triple product $\vec{a}$ . $\vec{b}$ * $\vec{c}$ given that $\vec{a}$ = $\vec{i}$ - $\vec{j}$ + 2 $\vec{k}$, $\vec{b}$ = 2 $\vec{i}$ + 3 $\vec{k}$ and $\vec{c}$ = 2 $\vec{i}$ - 4 $\vec{j}$ + $\vec{k}$

**Solution:**

Given $\vec{a}$ = $\vec{i}$ - $\vec{j}$ + 2 $\vec{k}$, $\vec{b}$ = 2 $\vec{i}$ +3 $\vec{k}$ and $\vec{c}$ = 2 $\vec{i}$ - 4 $\vec{j}$ + $\vec{k}$

$\vec{a}.\vec{b}*\vec{c} = \begin{vmatrix}

1 & -1& 2\\

2& 0 &3 \\

2 & -4 & 1

\end{vmatrix}$

= 1 (12) - 1 (4) + 2 (-8)

= 12 - 4 - 16

= - 8

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

The scalar product of two vectors satisfies the following properties.

- $\vec{a}$ . $\vec{b}$ = $\vec{b}$ . $\vec{a}$
- If $\vec{a}$ and $\vec{b}$ are both non zero vectors then $\vec{a}$ . $\vec{b}$ = 0 $\Rightarrow$ $\vec{a}$ and $\vec{b}$ are at right angles.
- If $\vec{a}$ and $\vec{b}$ are collinear vectors, then $\vec{a}$ . $\vec{b}$ = $\pm$ |$\vec{a}$| |$\vec{b}$|
- (m$\vec{a}$) . $\vec{b}$ = m ($\vec{a}$ . $\vec{b}$), if m is a scalar
- If $\vec{a}$ is any vector then $\vec{a}$ . $\vec{a}$ = |$\vec{a}$|$^{2}$
- If $\vec{a}$ and $\vec{b}$ are any two vectors and $\theta$ is the angle between them, then cos $\theta$ = $\frac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}$
- Dot product is distributive over vector addition.

### Solved Examples

**Question 1:**Add the vectors a = (9, 4) and b = (15, 8)

**Solution:**

Given : a = (9, 4) and b = (15, 8)

let c = a + b

$\rightarrow$ c = (9, 4) + (15, 8)

c = (9 + 15, 4 + 8)

c = (24, 12)

**Question 2:**Find $\vec{a}$ if 3 $\vec{a}$ + (- 2, 1, 5) = 2 $\vec{a}$ - (1, 0, 1)

**Solution:**

3 $\vec{a}$ + (- 2$\vec{i}$, 1$\vec{j}$, 5$\vec{k}$) = 2 $\vec{a}$ - (1$\vec{i}$, 0$\vec{j}$, 1$\vec{k}$)

3 $\vec{a}$ - 2 $\vec{a}$ = - ( - 2$\vec{i}$, 1$\vec{j}$, 5$\vec{k}$) - (1$\vec{a}$, 0$\vec{j}$, 1$\vec{k}$)

$\vec{a}$ = $\vec{i}$ - $\vec{j}$ - 6$\vec{k}$