Question

A point “P” is located in three dimensional coordinate plane such that the angles made by OP with positive direction of X - axis is ${30}^{\circ }$ and Y- axis is ${60}^{\circ }$ then find the angle made by OP with positive direction of Z - axis, where “O” is origin.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$

We know that cos of the angles made by a line segment with the positive direction of axes give us the direction cosines. Let the angle made with Z - axis be $\gamma$ , where 0 $\le \gamma \le \pi$
So, the direction cosines of OP will be $cos{30}^{\circ }$ ,$cos{60}^{\circ }$,$cos\gamma$
We know one property of direction cosines.
if the direction cosines of a line are l, m and n then
$l^2+m^2+n^2=1$
So,${\left(cos{30}^{\circ }\right)}^2+{\left(cos{60}^{\circ }\right)}^2+{\left(cos\gamma \right)}^2=1$
Or,${\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(cos\gamma \right)}^2=1$
Or,${\left(cos\gamma \right)}^2=0$
Or $cos\gamma =0$
Since $0\le \gamma \le \pi$ the only possible value $\gamma$ can have is $\frac{\pi }{2}$ radians.
So, $\gamma ={90}^{\circ }$