If $\mathbf{a}$ and $\mathbf{b}$ are two unit vectors such that $\mathbf{c}=(\mathbf{a} \times…
- 1
- $\frac{1}{2}$
- $\frac{3}{2}$
- 2
Solution

From Eq. (i) and (ii) $ \begin{aligned} & \frac{1}{2}\left\{1-|\mathbf{c}|^2 \cos ^2 \theta\right\}=|\mathbf{c}|^2 \sin ^2 \theta \Rightarrow|\mathbf{c}|^2=\frac{1}{\cos ^2 \theta+2 \sin ^2 \theta} \\ & \because\left[\begin{array}{lll} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{array}\right]=|\mathbf{c}|^2 \sin ^2 \theta=\frac{\sin ^2 \theta}{\cos ^2 \theta+2 \sin ^2 \theta}=\frac{\sin ^2 \theta}{1+\sin ^2 \theta} \\ & =\frac{1}{\operatorname{cosec}^2 \theta+1} \\ & \end{aligned} $ For maximum value of $\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right], \operatorname{cosec}^2 \theta=1$, so $\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]_{\max }=\frac{1}{2}$
Asked in: AP EAMCET 2018 (22 Apr Shift 2)