Question

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

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

Answer 1

The correct option is A $2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$

Let the angle subtended with each of the coordinate axis be Q

then

$l=m=n=\cos \theta$

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

${\cos }^2\theta +{\cos }^2\theta +{\cos }^2\theta =1$

$3{\cos }^2\theta =1$

${\cos }^2=\frac{1}{3}$

$\cos {\theta }^{}=\pm \frac{1}{\sqrt{3}}$

therefore l = m = n = $\pm \frac{1}{\sqrt{3}}$

then vector

$\begin{array}{c}\overrightarrow{v}=6\times \left(\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}\right)\end{array}$

$\begin{array}{c}\frac{6\sqrt{3}}{3}(\hat{i}+\hat{j}+\hat{k})\\ 2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})\end{array}$