91影视

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.

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

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

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

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

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

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

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