Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

We know that ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$ It is given that $\alpha +\beta =90^{\circ }\Rightarrow \alpha =90^{\circ }-\beta \Rightarrow \cos \alpha =\cos \left(90^{\circ }-\beta \right)$ $\therefore \cos \alpha =\sin \beta \Rightarrow {\cos }^2\alpha ={\sin }^2\beta =1-{\cos }^2\beta \Rightarrow {\cos }^2\alpha +{\cos }^2\beta =1$ Thus $1+{\cos }^2\gamma =1\Rightarrow {\cos }^2\gamma =0\Rightarrow \gamma =90^{\circ }$