Question

Show that the angle between any two diagonals of the cube is ${{\cos }^{-1}}\left( \dfrac{1}{3} \right)$.

Answer 1

Hint: To prove the angle between any two diagonals of the cube is ${{\cos }^{-1}}\left( \dfrac{1}{3} \right)$, first we have to draw a diagram of the cube. After drawing the cube, we have to mark the coordinates and mark the points. To solve this, we have to use the equation for the position vector. As the question is to prove the angle between any two diagonals of a cube is ${{\cos }^{-1}}\left( \dfrac{1}{3} \right)$, we have to use the equation to find the angle in the vector space.

Complete step-by-step answer:
First of all, we have to draw a diagram of the cube. We assume that the sides of the cube is ’a’. After that we have to mark the points and the corresponding coordinates. The origin is o (0, 0, 0). The cube is drawn in three dimensional form. So, the diagram will be,

The coordinates of a, b and c are (a, 0, 0), (0, a, 0) and (0, 0, a).
In the similar way the coordinates of d, e and g are (0, a, a), (a, 0, a) and (a, a, 0).
The four diagonals of the cube are ‘of’, ‘ad’, ‘be’ and ’cg’.
So, now we can consider the diagonals ‘ad’ and ‘be’.
Now, we have to find the position vector of these diagonals.
So, $\overrightarrow{ad}=$position vector of d – position vector of a = $(a\hat{j}+a\hat{k})-a\hat{i}=-a\hat{i}+a\hat{j}+a\hat{k}$
Similarly, $\overrightarrow{be}=$ position vector of e – position vector of b = $(a\hat{i}+a\hat{k})-a\hat{j}=a\hat{i}-a\hat{j}+a\hat{k}$
Now, we can assume that the angle between the diagonal is $\theta$.
Therefore, $\cos \theta =\dfrac{\overrightarrow{ad}\overrightarrow{be}}{\left| \overrightarrow{ad} \right|\left| \overrightarrow{be} \right|}$
On substituting the value we get,
$\cos \theta =\dfrac{(-a)(a)+(a)(-a)+(a)(a)}{\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}\times \sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}}$
On solving, we get,
$\cos \theta =\dfrac{-{{a}^{2}}-{{a}^{2}}+{{a}^{2}}}{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}$
$\begin{align}
  & \Rightarrow \cos \theta =\dfrac{-{{a}^{2}}}{3{{a}^{2}}} \\
 & \Rightarrow \cos \theta =\dfrac{-1}{3} \\
 & \therefore \theta ={{\cos }^{-1}}\left( \dfrac{-1}{3} \right) \\
\end{align}$

Note: The cube is drawn in the three dimension way. Any two diagonals can be taken and solved. The position vector of $\overrightarrow{ad}=$position vector of d – position vector of a. The angle between any two diagonal can be found using the equation $\cos \theta =\dfrac{\overrightarrow{ad}\overrightarrow{be}}{\left| \overrightarrow{ad} \right|\left| \overrightarrow{be} \right|}$.