Two ropes are tied along the two diagonals of a cubic room. Ropes are inclined to each other at :

$\cos ^{-1}\left(\frac{1}{3}\right)$

Let us imagine a cube of unit distances

→  one diagonal is along

HE → H(0, 0, 0)    E(1, 1, 1)

one along ac

→ a(0, 1, 1) to c(1, 0, 0)

So $\vec{HE} = \hat{i} + \hat{j} + \hat{k} = \vec{v_1}$

$\vec{ac} = \hat{i} - \hat{j} - \hat{k} = \vec{v_2}$

angle between ropes

→  $\vec{v_1} . \vec{v_2} = |\vec{v_1}| |\vec{v_2}| \cos \theta$

⇒  $|1 - 1 - 1| = \sqrt{3} \sqrt{3} \cos \theta$

$\cos \theta = \frac{1}{3} \Rightarrow  \theta = \cos ^{-1}\left(\frac{1}{3}\right)$