# Multiplication of a Vector by a Scalar

## Multiplication of a Vector by a Scalar

Let $$\overline{a}$$ be a vector and λ a scalar. Then the product of vector $$\overline{a}$$ by scalar λ, denoted by $$\lambda \overline{a}$$, is called the multiplication of vector  by the scalar λ.

Note that $$\lambda \overline{a}$$ is also a vector, collinear to vector $$\overline{a}$$. Vector $$\lambda \overline{a}$$ has the direction same (or opposite) as that of vector $$\overline{a}$$ if the value of λ is positive (or negative). Also, the magnitude of vector $$\lambda \overline{a}$$ is |λ| time the magnitude of vector  $$\overline{a}$$ , i. e.,

$$\left| \lambda \overline{a} \right|=\left| \lambda \right|\left| \overline{a} \right|$$.

A geometric visualization of multiplication of a vector by a scalar is given in the following figure.

When $$\lambda =-1,\ \lambda \overline{a}=-\overline{a}$$, which is a vector having magnitude equal to the magnitude of $$\overline{a}$$ and direction opposite to that of the direction of $$\overline{a}$$.

Vector $$-\overline{a}$$ is called the negative (or additive inverse) of vector $$\overline{a}$$ .

$$\overline{a}+(-\overline{a})=\left( \overline{-a} \right)+\overline{a}=0$$.

Also, if $$\lambda =\frac{1}{\left| \overline{a} \right|}$$, provided $$\overline{a}\ne 0$$ i.e., $$\overline{a}$$ is not a null vector, then

$$\left| \lambda \overline{a} \right|=\left| \lambda \right|\left| \overline{a} \right|=\frac{1}{\left| \overline{a} \right|}\times \left| \overline{a} \right|=1$$.

Example: Find a vector in the direction of vector $$5\overline{i}-\overline{j}+2\overline{k}$$ which has magnitude 8 units

Solution:

Given that $$5\overline{i}-\overline{j}+2\overline{k}$$,

Let $$\overline{a}=5\overline{i}-\overline{j}+2\overline{k}$$,

$$\left| \overline{a} \right|=\sqrt{{{5}^{2}}+{{(-1)}^{2}}+{{2}^{2}}}$$,

$$=\sqrt{25+1+4}=\sqrt{30}$$.

$$\widehat{a}=\frac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$$,

$$=\frac{5\overline{i}-\overline{j}+2\overline{k}}{\sqrt{30}}$$.

Hence, the vector in the direction of vector $$5\overline{i}-\overline{j}+2\overline{k}$$ which has magnitude 8 units is given by

$$8\widehat{a}=8\left( \frac{5\overline{i}-\overline{j}+2\overline{k}}{\sqrt{30}} \right)$$,

$$=\frac{40}{\sqrt{30}}\overline{i}-\frac{8}{\sqrt{30}}\overline{j}+\frac{16}{\sqrt{30}}\overline{k}$$.