Angle between internal diagonal and a edge of a cuboid

Question

AP=DS=CR=BQ=2a, ABCD and PQRS are squares with a side a find the angle between AR and BC.. i got $$AR=\sqrt{6}a$$can i use ARQ triangle in order to find the corresponding angle..

Answer 1

The triangle $\Delta ACR$ has a right angle at $C$. The side $AC = \sqrt{2} CR$, so $\tan \alpha = \frac{1}{\sqrt{2}}$, and you get your $\alpha = \arctan \frac{1}{\sqrt{2}}= \ldots$.

Btw, the large diagonal is $\sqrt{3} \cdot \textrm{edge} = 2\sqrt{3} a$, if you want to use $\arcsin$.

Answer 2

Take $R$ as the origin of vectors. Then $\vec{AR}$ is along $a\mathbf{i} + a\mathbf{j} +2a\mathbf{k}$ and $\vec{CB}$ is along $a\mathbf{i}$. The angle between them is hence $$\cos^{-1}\left( \frac{(\mathbf{i} + \mathbf{j} +2\mathbf{k})\cdot \mathbf{i}}{\sqrt{6}}\right) = \cos^{-1}\frac{1}{\sqrt{6}}$$