Question

A vector $\overrightarrow{r}$ is inclined at equal angles to $OX,OY$ and $OZ$. If the magnitude of $r$ is $6$ units, then $\overrightarrow{r}$ is equal to

• A. $\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$
• B. $-\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$
• C. $2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$
• D. $-2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$

Answer 1

Answer: (C), (D)

The correct options are
C $2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$
D $-2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$

Let $r$ be inclined at an angle $\alpha$ to each axis, then $l=m=n=\cos \alpha$

Since $l^2+m^2+n^2=1$

$\Rightarrow 3{\cos }^2\alpha =1$

$\Rightarrow \cos \alpha =\pm \frac{1}{\sqrt{3}}$

If $\alpha$ is acute: $l=m=n=\frac{1}{\sqrt{3}}:\left|r\right|=6$

$r=\left|r\right|(l\hat{i}+m\hat{i}+n\hat{k})=6(\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k})=2\sqrt{3}(\hat{i}+\hat{j}+\hat{k}).$

If $\alpha$ is obtuse $l=m=n=-\frac{1}{\sqrt{3}}:\left|r\right|=6$

$r=\left|r\right|(l\hat{i}+m\hat{j}+n\hat{k})=6(-\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k})$

$=-2\sqrt{3}(\hat{i}+\hat{j}+\hat{k}).$