In linear algebra, real numbers are called scalars. The word scalar is used to mean a vector, matrix that is reduced to a single component. Scalar is a positive real number that multiplies the magnitude of the vector without changing its direction. |

Then the scalar product of a and b is defined as

k(a . b) = k.( |a| |b| cos $\theta$)

Where $\theta$ is the angle between a and b.

There is also called

**cross product**in vectors, which has both magnitude as well as direction.

We use a large diagonal cross (*) symbol to represent cross product. It is also called vector product.

Considers two vectors $\vec{a}$ and $\vec{b}$ with a scalar k then their cross product is defined as

k . |$\vec{a} * \vec{b}$| = k . |$\vec{a}$||$\vec{b}$| sin $\theta$ $\hat{n}$

**For vectors u, v, w and scalar a, b, c the following holds :**

Left Distributive property: a(u + v) = au + av

Right Distributive property: (a + b)u = au + bu

Associative property: a(bu) = (ab)(u) = (ba)(u)

Commutative property: au = ua

Identity Property: 1.u = u

Multiplicative property of 0: 0.(u) = 0

Multiplicative property of -1: (-1) a = - a

Magnitude property of a scalar: |au| = |a|u Scalar multiplication is easy to apply in vectors. By performing the required operations on the given vector components we can easily do the scalar multiplication. Given below is an example to illustrate the concept of scalar multiplication.

Let vectors A and B are given as A = 3i + 2j + k and B = - i + 4j + 9k. To find 10(AB) we need to follow the below procedure.

For the given problem, first find the dot product for both the vectors then multiply the result by 10.

Here A = 3i + 2j + k

B = - i + 4j + 9k

A . B = (3i + 2j + k) . ( - i + 4j + 9k)

= 3(-1) + 2 (4) + 1 (9)

= - 3 + 8 + 9

A . B = 14

Now, 10(A.B) = 10(14)

= 140

Therefore the value of 10(AB) = 140.

**Example 1:**Let k = ( 1, 2, -5). What will be the value of 7k?

**Solution:**Given k = (1, 2, -5)

Now to find 7k, multiply k with 7.

7k = 7(1, 2, -5)

= (7, 14, -35)

Therefore the value of 7k is (7, 14, -35).

**Example 2:**Let a = (2, -3, 4) and b = (-1, 4, 3). Find the value of 11(axb).

**Solution:**Given a = (2, -3, 4)

b = (-1, 4, 3)

To find the value of 11(axb) we first need to find axb

a x b = $\begin{vmatrix}

i &j &k\\

2 & -3 & 4\\

-1 & 4& 3

\end{vmatrix}$

= i(-3.3 - 4.4) -j(2.3 +1.4) + k(2.4 -1.3)

= -25i - 10j + 5k

Now, 11(axb) = 11(-25i -10j + 5k)

= - 275i - 110j + 55k.

Therefore 11 (axb ) = - 275i - 110j + 55k.

**Example 3:**Let $\vec{a}$ = (-1, 2, 3) and $\vec{b }$ = (-2, 4, -5). Find 5a * b

**Solution:**Given $\vec{a}$ = (-1, 2, 3)

$\vec{b}$ = (-2, 4, -5)

To find 5a * b, first find 5a.

5a = 5(-1, 2, 3)

= (- 5, 10, 15)

5a * b = $\begin{vmatrix}

i &j &k\\

-5 & 10 &15\\

-2 & 4& - 5

\end{vmatrix}$

= i (10 . -5 - 15 . 4) - j (-5 . -5 + 2 . 15) + k ( -5 . 4 + 2 . 10)

= -110i - 55j + 0k

Therefore, 5a * b = (-110i - 55j)