Question

The angle between two diagonals of a cube is

• A. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• B. $30°$
• C. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. $45°$

Answer 1

The correct option is A

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

Explanation for the correct option:

Finding the angle between two diagonals of a cube

We know that $\cos \theta =\frac{\mid u.v\mid }{\mid u\mid \mid v\mid }$
Consider a cube with one vertex at the origin.
Then one diagonal is from

$(0,0,0)\text{to}(1,1,1):\mid \overrightarrow{u}\mid =<1,1,1>$
Another diagonal is from

$(1,0,0)\mathrm{to}(0,1,1):\mid \overrightarrow{v}\mid =<1,-1,-1>$
Therefore,

$\cos \theta =\frac{\left|<1,1,1>.<1,-1,-1>\right|}{\left|\sqrt{1^2+1^2+1^2}\right|\left|\sqrt{1^2+{(-1)}^2+{(-1)}^2}\right|}$
$$\begin{gathered} \cos \theta =\frac{\left|1-1-1\right|}{\sqrt{3}\sqrt{3}} \\ \cos \theta =\frac{1}{3} \end{gathered}$$
Therefore, $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$

Hence, the correct answer is an option (A).