Question

let $\overrightarrow{\alpha }=a\hat{i}+b\hat{j}+c\hat{k}$ and $\overrightarrow{\beta }=b\hat{i}+c\hat{j}+a\hat{k}$ where a,b,c $\in R$ . If $\theta$ is the angle between the two vectors then

• A. $\theta$ $\in$ (0,$\frac{\pi }{2}$)
• B. $\theta$ $\in$ [0,$\frac{2\pi }{3}$]
• C. $\theta$ $\in$ ($\frac{2\pi }{3}$,$\pi$ ]
• D. none of these

Answer 1

The correct option is B $\theta$ $\in$ [0,$\frac{2\pi }{3}$]

$\begin{array}{c}\cos \theta =\frac{ab+bc+ca}{a^2+b^2+c^2}\\ \\ and,\\ {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\ge 0\\ a^2+b^2+c^2\ge ab+bc+ca\\ \frac{ab+bc+ca}{a^2+b^2+c^2}\le 1\\ \\ and,\\ {\left(a+b+c\right)}^2\ge 0\\ a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge 0\\ ab+bc+ca\ge \frac{-\left(a^2+b^2+c^2\right)}{2}\\ \frac{ab+bc+ca}{a^2+b^2+c^2}\ge \frac{-1}{2}\\ \\ then,\\ \cos \theta \in \left[\frac{-1}{2},1\right]\\ \theta \in \left[0,\frac{2\pi }{3}\right]\end{array}$