The cosines of the angle between any two diagonals of a cube is

The correct option is 1/3.

Let us find the cosine of the angle between any two diagonals of a cube step-by-step using vector algebra.

Consider a cube with one of its vertices placed at the origin of a three-dimensional Cartesian coordinate system. Let the three mutually perpendicular edges of the cube lie along the positive coordinate axes, and let the length of each edge of the cube be $a$.

The vertices of this cube can be represented by the following coordinates:
Origin $O(0,0,0)$
Vertices on the axes: $A(a,0,0)$, $B(0,a,0)$, $C(0,0,a)$
Other vertices: $D(a,a,0)$, $E(0,a,a)$, $F(a,0,a)$, and the opposite corner to the origin $G(a,a,a)$.

The four body diagonals of the cube connect the opposite pairs of vertices:
1. From $O(0,0,0)$ to $G(a,a,a)$
2. From $A(a,0,0)$ to $E(0,a,a)$
3. From $B(0,a,0)$ to $F(a,0,a)$
4. From $C(0,0,a)$ to $D(a,a,0)$

Let us choose two of these diagonals to find the angle between them. We will choose diagonal $OG$ and diagonal $AE$.

Let $\hat{i}$, $\hat{j}$, and $\hat{k}$ be the unit vectors along the positive x-axis, y-axis, and z-axis, respectively.

The vector representing the diagonal $OG$ is given by:
$\overrightarrow{d_1}=(a-0)\hat{i}+(a-0)\hat{j}+(a-0)\hat{k}=a\hat{i}+a\hat{j}+a\hat{k}$

The vector representing the diagonal $AE$ is given by:
$\overrightarrow{d_2}=(0-a)\hat{i}+(a-0)\hat{j}+(a-0)\hat{k}=-a\hat{i}+a\hat{j}+a\hat{k}$

The angle $\theta$ between two vectors is given by the dot product formula:
$\cos \theta =\frac{\overrightarrow{d_1}\cdot \overrightarrow{d_2}}{\left|\overrightarrow{d_1}\right|\left|\overrightarrow{d_2}\right|}$

First, we calculate the dot product of the two vectors:
$\overrightarrow{d_1}\cdot \overrightarrow{d_2}=\left(a\right)(-a)+\left(a\right)\left(a\right)+\left(a\right)\left(a\right)$
$\overrightarrow{d_1}\cdot \overrightarrow{d_2}=-a^2+a^2+a^2=a^2$

Next, we calculate the magnitude of each vector:
$\left|\overrightarrow{d_1}\right|=\sqrt{a^2+a^2+a^2}=\sqrt{3a^2}=a\sqrt{3}$
$\left|\overrightarrow{d_2}\right|=\sqrt{{(-a)}^2+a^2+a^2}=\sqrt{3a^2}=a\sqrt{3}$

Now, substitute these values back into the cosine formula:
$\cos \theta =\frac{a^2}{\left(a\sqrt{3}\right)\left(a\sqrt{3}\right)}$
$\cos \theta =\frac{a^2}{3a^2}=\frac{1}{3}$

Thus, the cosine of the angle between any two diagonals of a cube is $\frac{1}{3}$.