Q. 41 - Calculus

Question

Find the angle between the diagonal of a cube and an adjoining diagonal of one of the faces of the cube.

Step-by-Step Solution

Answer

The angle between the diagonal of a cube and an adjoining diagonal of one of the faces of the cube is \(cos^{-1}\left ( \frac{\sqrt{2}}{\sqrt{3}} \right )\).

Step 1. Given Information

An angle can be defined as a space formed when 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 diagonal of a cube and an adjoining diagonal of one of the faces of the 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, EG and DE.

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

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

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

And

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

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

Now, the magnitude of the vectors are 

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

\(\left\|\underset{EG}{\rightarrow} \right\|=\sqrt{3}\)

And

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

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

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, 

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

\(\underset{EG}{\rightarrow}\cdot \underset{DE}{\rightarrow}=2\)

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

\(\theta =cos^{-1}\left ( \frac{2}{\sqrt{6}} \right )\) 

\(\theta =cos^{-1}\left ( \frac{\sqrt{2}}{\sqrt{3}} \right )\)