Step 1: Represent the cube using vectors.
Consider a cube of side length $a$. Take one edge of the cube along the $x$-axis, represented by the vector
\[
\vec{e} = (a, 0, 0).
\]
The longest diagonal of the cube connects opposite vertices and is represented by the vector
\[
\vec{d} = (a, a, a).
\]
Step 2: Use the dot product formula.
The cosine of the angle $\theta$ between two vectors $\vec{e}$ and $\vec{d}$ is given by
\[
\cos \theta = \frac{\vec{e} \cdot \vec{d}}{|\vec{e}|\,|\vec{d}|}.
\]
Step 3: Compute dot product and magnitudes.
\[
\vec{e} \cdot \vec{d} = a^2,
|\vec{e}| = a,
|\vec{d}| = a\sqrt{3}.
\]
Step 4: Evaluate $\cos \theta$.
\[
\cos \theta = \frac{a^2}{a \cdot a\sqrt{3}} = \frac{1}{\sqrt{3}}.
\]
Step 5: Conclusion.
Hence, the correct value of $\cos \theta$ is $\dfrac{1}{\sqrt{3}}$.

In \(\triangle ABC\), \(DE \parallel BC\). If \(AE = (2x+1)\) cm, \(EC = 4\) cm, \(AD = (x+1)\) cm and \(DB = 3\) cm, then the value of \(x\) is

In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90^\circ$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is
In the adjoining figure, TS is a tangent to a circle with centre O. The value of $2x^\circ$ is