Find the position vector of a point A in space such that OA is inclin…

The correct answer is Option (1) → $5\hat{i} + 5\sqrt{2}\hat{j} \pm 5\hat{k}$ ##

Given, $OA$ is inclined at $60^\circ$ to $OX$ and at $45^\circ$ to $OY$. Let $OA$ makes angle $\alpha$ with $OZ$.

Then $\cos^2 60^\circ + \cos^2 45^\circ + \cos^2 \alpha = 1 \quad [∵l^2 + m^2 + n^2 = 1]$

$\Rightarrow \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \alpha = 1$

$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \alpha = 1$

$\Rightarrow \cos^2 \alpha = 1 - \left(\frac{1}{2} + \frac{1}{4}\right)$

$\Rightarrow \cos^2 \alpha = 1 - \left(\frac{3}{4}\right)$

$\Rightarrow \cos^2 \alpha = \frac{1}{4}$

$\Rightarrow \cos \alpha = \frac{1}{2} = \cos 60^\circ$

$∴\alpha = 60^\circ$

Now, $\vec{OA} = |OA| \left( \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k} \right)$

$= 10 \left( \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k} \right) \quad [∵|OA| = 10]$

$= 5 \hat{i} + 5\sqrt{2} \hat{j} + 5 \hat{k}$