Consider a pyramid OPQRS located in the first octant (x≥0, y ≥0, z ≥0) with O as origin, and OP and OR along the x-axis and y-axis respectively. The base OPQR of the pyramid is a square with OP = 3. The point S is directly above the mid-point T of the diagonal OQ such that TS = 3. Then, the angle between OQ and OS, is

$\cos^{-1}\frac{1}{\sqrt{3}}$

The coordinates of T and S are (3/2, 3/2, 0) and (3/2, 3/2, 3) respectively.

$∴\vec{OQ}=3\hat i+ 3\hat j$ and $\vec{OS} = \frac{3}{2}\hat i+\frac{3}{2}\hat j+ 3\hat k$

Let θ be the angle between $\vec{OQ}$ and $\vec{OS}$. Then,

$\cos θ=\frac{\vec{OQ}.\vec{OS}}{|\vec{OQ}||\vec{OS}|}$

$⇒\cos θ=\frac{\frac{9}{2}+\frac{9}{2}+0}{\sqrt{9+9}\sqrt{\frac{9}{4}+\frac{9}{4}+9}}$

$⇒\cos θ=\frac{9}{3\sqrt{2}×\sqrt{\frac{27}{2}}}=\frac{1}{\sqrt{3}}$