Question

In a unit cube. Find The angle between the diagonal of a cube and a diagonal of a face intersecting it ?

Answer 1

Here, CD is a diagonal and CA is a face,

so, diagonal intersecting it,

D.r's of CD is 1, 1, -1

D.r's of CA is 1, 0, 1

D.c's of CD is $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}}$

D.c'r of ca is $\frac{1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}}$

Angle between them is

$\begin{array}{c}\cos \theta =\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{2}}+0+\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{2}}\\ =\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}}\left[\therefore \sqrt{2}=1\right]\\ so,\\ =\sqrt{\frac{4}{6}}=\sqrt{\frac{2}{3}}\\ \therefore \theta ={\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)\end{array}$