If $\mathbf{a}$ and $\mathbf{b}$ are two unit vectors such that $\mathbf{c}=(\mathbf{a} \times \mathbf{c})+\mathbf{b}$, then the maximum value of $[\mathbf{a} \mathbf{b c}]$ is

• A

  1
• Correct answer
  B

  $\frac{1}{2}$
• C

  $\frac{3}{2}$
• D

  2

Solution

$$ \begin{aligned} & \mathbf{c}=(\mathbf{a} \times \mathbf{c})+\mathbf{b} \\ & \Rightarrow \quad|\mathbf{c}|^2=|\mathbf{a}+\mathbf{c}|^2+|\mathbf{b}|^2+2(\mathbf{a} \times \mathbf{c}) \cdot \mathbf{b} \\ & \Rightarrow \quad 2[\mathbf{a b c}]=|\mathbf{a} \times \mathbf{c}|^2+1-|\mathbf{c}|^2 \end{aligned} $$ Let angle between vectors $\mathbf{a}$ and $\mathbf{c}$ is ' $\theta$ '. So, $\quad\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]=\frac{1}{2}\left[|\mathbf{c}|^2 \sin ^2 \theta-|\mathbf{c}|^2+1\right]$
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 $\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]_{\max }=\frac{1}{2}$

Hence, the correct answer is option: 'B'.