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,

$\underset{AG}{\to }=⟨1-0,0-0,1-0⟩$

$\underset{AG}{\to }=⟨1,0,1⟩$

And

$\underset{DF}{\to }=⟨0-0,0-1,1-0⟩$

$\underset{DF}{\to }=⟨0,-1,1⟩$

Now, the magnitude of the vectors are

$\Vert \underset{AG}{\to }\Vert =\sqrt{1^2+0^2+1^2}$

$\Vert \underset{AG}{\to }\Vert =\sqrt{2}$

And

$\Vert \underset{DF}{\to }\Vert =\sqrt{0^2+(-1{)}^2+1^2}$

$\Vert \underset{DF}{\to }\Vert =\sqrt{2}$

Step 3. Find the angle

Now, the angle between two vectors $\underset{a}{\to }$ and $\underset{b}{\to }$ is $\theta =co{s}^{-1}\frac{\underset{a}{\to }\cdot \underset{b}{\to }}{\left|\underset{a}{\to }\right|\left|\underset{b}{\to }\right|}$ .

So,

$\underset{AG}{\to }\cdot \underset{GF}{\to }=⟨1,0,1⟩\cdot ⟨0,-1,1⟩$

$\underset{AG}{\to }\cdot \underset{GF}{\to }=1$

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

$\theta =co{s}^{-1}\frac{1}{\sqrt{2}\sqrt{2}}$

$\theta =co{s}^{-1}\left(\frac{1}{2}\right)$

$\theta =\frac{\pi }{6}$