Question

Cosine of the angle between two diagonals of a cube is equal to

• A. $\frac{2}{\sqrt{6}}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{3}$

Let 0, one vertex of a cube, be the origin and three edges through O be the coordinate axes. The four diagonals are Op, AA', BB' and CC'. Let a be the length of each edge. Then the coordinates of P, A, A' are (a, a, a), (a, 0, 0), (0, a, a).
The direction ratios of OP are a, a, a.
The direction cosines of OP are
$\frac{a}{a\sqrt{3}},\frac{a}{a\sqrt{3}},\frac{a}{a\sqrt{3}}$ i.e., $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.
Similarly direction cosines of AA'are
$(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$.
Let $\theta$ be the angle between the diagonals OP and AA'.
$cos\theta =\frac{1}{\sqrt{3}}(-\frac{1}{\sqrt{3}})+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)+\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}}\right)$
$cos\theta =l_1{l}_2+m_1{m}_2+n_1{n}_2=-\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{3}$
$\therefore \theta =co{s}^{-1}\left(\frac{1}{3}\right)$.