Question

Prove Theorem 10.11; that is, show that when $v\ne 0$, the scaled vector role="math" localid="1663644167280" $\frac{1}{\left|v\right|}v$ is a unit vector with the same direction as $v$.

Short answer

It is proven that,$\frac{1}{\left|v\right|}v$is a unit vector with the same direction as$v$.

Step-by-step solution

Consider a vector

Let us consider that,

$v=i+j+k$

Proof

Now calculate the magnitude of vector $v$,

$$\begin{gathered} \left|v\right|=\sqrt{1^2+1^2+1^2} \\ =\sqrt{1+1+1} \\ =\sqrt{3} \end{gathered}$$

The scaled vector $\frac{1}{\left|v\right|}v$is,

$$\begin{gathered} \frac{1}{\left|v\right|}v=\frac{1}{\sqrt{3}}\left(i+j+k\right) \\ =\frac{1}{\sqrt{3}}i+\frac{1}{\sqrt{3}}j+\frac{1}{\sqrt{3}}k \end{gathered}$$

The magnitude of the scaled vector $\frac{1}{\left|v\right|}v$is computed as,

$$\begin{gathered} \left|\frac{1}{\left|v\right|}v\right|=\sqrt{{\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2} \\ =\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}} \\ =\sqrt{\frac{3}{3}} \\ =\sqrt{1} \\ =1 \end{gathered}$$

Since the magnitude of the scaled vector is unity.

Therefore, the scaled vector is a unit vector.

Hence, proved.