Question

Find the position vector of a point A in space such that $\overrightarrow{OA}$ is inclined at ${60}^{\circ }$ to OX and at ${45}^{\circ }$ to OY and $\overrightarrow{|OA|}$= 10 units.

Answer 1

Since, $\overrightarrow{OA}$ is inclined at ${60}^{\circ }$ to OX and ${45}^{\circ }$ to OY. Let $\overrightarrow{OA}$ makes angle $\alpha$ with OZ.
$\therefore co{s}^2{60}^{\circ }+co{s}^2{45}^{\circ }+co{s}^2\alpha =1\Rightarrow {\left(\frac{1}{2}\right)}^2+\Rightarrow {\left(\frac{1}{\sqrt{2}}\right)}^2+co{s}^2\alpha =1[\because l^2+m^2+n^2=1]\Rightarrow \frac{1}{4}+\frac{1}{2}+co{s}^2\alpha =1\Rightarrow co{s}^2\alpha =1-(\frac{1}{2}+\frac{1}{4})\Rightarrow co{s}^2\alpha =1-\left(\frac{6}{8}\right)\Rightarrow co{s}^2\alpha =\frac{1}{4}\Rightarrow cos\alpha =\frac{1}{2}=cos{60}^{\circ }\therefore \alpha ={60}^{\circ }\therefore \overrightarrow{OA}=\overrightarrow{|OA|}(\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k})=10(\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k})[\because \overrightarrow{|OA|}=10]=5\hat{i}+5\sqrt{2}\hat{j}+5\hat{k}$