Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Consider a cube of side length $a$ with one of its vertices at the origin $(0,0,0)$ and edges along the coordinate axes.

The vector representing an edge along the x-axis is $\vec{e}=a\hat{i}$.

The vector representing the diagonal of the cube from the origin is $\vec{d}=a\hat{i}+a\hat{j}+a\hat{k}$.

The angle $\theta$ between the diagonal and the edge is given by $\cos \theta =\frac{\vec{d}\cdot \vec{e}}{|\vec{d}||\vec{e}|}$.

$\vec{d}\cdot \vec{e}=(a)(a)+(a)(0)+(a)(0)=a^2$

$|\vec{d}|=\sqrt{a^2+a^2+a^2}=a\sqrt{3}$

$|\vec{e}|=a$

$\cos \theta =\frac{a^2}{a\sqrt{3}\times a}=\frac{1}{\sqrt{3}}$

Using trigonometric identities, if $\cos \theta =\frac{1}{\sqrt{3}}$, then the adjacent side is $1$ and the hypotenuse is $\sqrt{3}$.

The opposite side is $\sqrt{(\sqrt{3}{)}^2-1^2}=\sqrt{2}$.

Therefore, $\tan \theta =\frac{\sqrt{2}}{1}=\sqrt{2}$.

$\theta ={\tan }^{-1}(\sqrt{2})$

Answer: ${\tan }^{-1}\left(\sqrt{2}\right)$