Question

Find the angle between two distinct diagonals of a cube.

Short answer

The angle between two distinct diagonals of a cube is \(\frac{\pi }{6}\).

Step-by-step solution

Step 1. Given Information

An angle can be defined as a space formedwhen two straight lines or rays meet at a common endpoint.

The diagonal of a cube can be defined as the line segment that joins any two non-adjacent vertices in it. In Cube, we have 12 face diagonals 2 on each face, and 4 space diagonals.

Step 2. Find the position vector of the diagonals

To find the angle between the two distinct diagonals of a cube. Let the coordinates of ABCDEFGH be \((0, 0, 0\)), \((1, 0, 0\)), \((1, 1, 0\)), \((0, 1, 0\)), \((0, 1, 1), (0, 0, 1\)), \((1, 0, 1\)), \((1, 1, 1\)).

Now, let's consider the two diagonals, AG and DF.

Let's find the position vector of these diagonals so,

\(\underset{AG}{\rightarrow}=\left<1-0, 0-0, 1-0 \right>\)

\(\underset{AG}{\rightarrow}=\left<1,0,1 \right>\)

And

\(\underset{DF}{\rightarrow}=\left<0-0, 0-1, 1-0 \right>\)

\(\underset{DF}{\rightarrow}=\left<0,-1,1 \right>\)

Now, the magnitude of the vectors are

\(\left\|\underset{AG}{\rightarrow} \right\|=\sqrt{1^{2}+0^{2}+1^{2}}\)

\(\left\|\underset{AG}{\rightarrow} \right\|=\sqrt{2}\)

And

\(\left\|\underset{DF}{\rightarrow} \right\|=\sqrt{0^{2}+(-1)^{2}+1^{2}}\)

\(\left\|\underset{DF}{\rightarrow} \right\|=\sqrt{2}\)

Step 3. Find the angle

Now, the angle between two vectors \(\underset{a}{\rightarrow}\) and \(\underset{b}{\rightarrow}\) is \(\theta=cos^{-1} \frac{\underset{a}{\rightarrow}\cdot \underset{b}{\rightarrow}}{\left| \underset{a}{\rightarrow}\right|\left|\underset{b}{\rightarrow} \right|}\) .

So,

\(\underset{AG}{\rightarrow}\cdot \underset{GF}{\rightarrow}=\left< 1,0,1\right>\cdot \left<0,-1,1 \right>\)

\(\underset{AG}{\rightarrow}\cdot \underset{GF}{\rightarrow}=1\)

Let's put all the values in the formula of the angle between two vectors

\(\theta=cos^{-1} \frac{1}{\sqrt{2}\sqrt{2}}\)

\(\theta =cos^{-1}\left ( \frac{1}{2} \right )\)

\(\theta =\frac{\pi }{6}\)