Maximum angle between two vectors obtained by cyclic permutations of coordinates

Question

If $a, b, c$ are direction cosines of a line and $\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}$, $\vec{B} = b\hat{i} + c\hat{j} + a\hat{k}$, then find the maximum angle $\theta$ (in degrees) between $\vec{A}$ & $\vec{B}$ where $\theta \in [0,\pi]$.

Answer 1

Since $a, b, c$ are direction cosines of a line, $a^2 + b^2 + c^2 = 1$

We know that $(a + b + c)^2 \ge 0$ $\implies$ $a^2 + b^2 + c^2 + 2(ab + bc + ca) \ge 0$ $\implies$ $ab+bc+ca \ge \frac{-1}{2}$

Now to calculate angle between $\vec{A}$ & $\vec{B}$ we should use dot product of two vectors.

$\implies$ $\vec{A}.\vec{B} = |\vec{A}||\vec{B}|\cos\theta$ $\implies$ $ab+bc+ca = (\sqrt{a^2 + b^2 + c^2})(\sqrt{a^2 + b^2 + c^2})\cos\theta$ $\implies$ $\cos\theta \ge \frac{-1}{2}$

Therefore the maximum possible value of $\theta \in [0,\pi]$ is $120^o$