Question

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

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. 2

Answer 1

Answer: (B)

$c=(a\times c)+b$
$\Rightarrow |c{|}^2=|a+c{|}^2+|b{|}^2+2(a\times c)\cdot b$
$\Rightarrow 2[abc]=|a\times c{|}^2+1-|c{|}^2$
Let angle between vectors a and c is ${}^{\prime }{\theta }^{\prime }$
So, $[abc]=\frac{1}{2}[|c{|}^2{\sin }^2\theta -|c{|}^2+1]$
$=\frac{1}{2}\left[1-|c{|}^2{\cos }^2\theta \right]$ ...(i)
Now, $c\cdot (a\cdot c)=(a\times c{)}^2+b\cdot (a\times c)$
$\Rightarrow [abc]=|c{|}^2{\sin }^2\theta$ ...(ii)
From Eq. (i) and (ii)
$\frac{1}{2}\left\{1-|c{|}^2{\cos }^2\theta \right\}=|c{|}^2{\sin }^2\theta$
$\Rightarrow |c{|}^2=\frac{1}{{\cos }^2\theta +2{\sin }^2\theta }$
$\because [abc]=|c{|}^2{\sin }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta +2{\sin }^2\theta }=\frac{{\sin }^2\theta }{1+{\sin }^2\theta }$
$=\frac{1}{{\mathrm{cosec}}^2\theta +1}$
For maximum value of $[abc],{\text{cosec}}^2\theta =1$,
so $[abc{]}_{\max }=\frac{1}{2}$