Question

Let the angle between two diagonals of a cube be $\arccos \frac{1}{k}$. Find $k$ ?

Answer 1

Let $O$ be one vertex of a cube, and three edge through $O$ be the coordinate axes.
The four diagonal are $OP,A{A}^{\prime },BB,$ and $C{C}^{\prime }$
Let $a$ be the length of each edge.
Then the coordinate of $P,A,A^{\prime }$ are $(a,a,a),(a,0,0),(0,a,a)$
The direction ratios of $OP$ are
$\frac{a}{a\sqrt{3}},\frac{a}{a\sqrt{3}},\frac{a}{a\sqrt{3}}\Rightarrow \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
Similarly direction cosine of $A{A}^{\prime }$ are
$(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$
Let $\theta$ be the angle between the diagonals $Op$ and $A{A}^{\prime }$, then
$\cos \theta =\frac{1}{\sqrt{3}}(-\frac{1}{\sqrt{3}})+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)$
$=-\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{3}$
$\therefore \theta ={\cos }^{-1}\frac{1}{3}$