Question

If $\overline{a}$ and $\overline{b}$ are units vectors and $\overline{c}$ satisfies $2\left(\overline{a}\times \overline{b}\right)+\overline{c}=\overline{b}\times \overline{c}$ then the maximum value of $\left|\left(\overline{a}\times \overline{c}\right).\overline{b}\right|$ is

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

Answer 1

Answer: (B)

$2$

$2(\overline{a}\times \overline{b})+\overline{c}=\overline{b}\times \overline{c}-\left(1\right)$

Doing dot product of L.H.S.& R.H.S. with $\overline{c}$

$\Rightarrow 2(\overline{a}\times \overline{b}).\overline{c}+\overline{c}.\overline{c}=(\overline{b}\times \overline{c}).\overline{c}$

$\Rightarrow 2(\overline{a}\times \overline{b}).\overline{c}+|\overline{c}{|}^2=0$

$\Rightarrow 2(\overline{a}\times \overline{b}).\overline{c}=-|\overline{c}{|}^2$

$\Rightarrow |(\overline{a}\times \overline{b}).\overline{c}|=\frac{|\overline{c}{|}^2}{2}-\left(2\right)$

Note : $|(\overline{a}\times \overline{c}).\overline{b}|=|(\overline{a}\times \overline{b}).\overline{c}|=|(\overline{b}\times \overline{c}).\overline{a}|$

$\left(1\right)\to 2(\overline{a}\times \overline{b})+\overline{c}=\overline{b}\times \overline{c}$

Doing dot product of both side with $\overline{a}$

$\Rightarrow 2(\overline{a}\times \overline{b}).\overline{a}+\overline{c}.\overline{a}=(\overline{b}\times \overline{c}).\overline{a}$

$\Rightarrow 0+\overline{c}.\overline{a}=(\overline{b}\times \overline{c}).\overline{a}$

$|\overline{a}.(\overline{b}\times \overline{c})|=|\overline{c}.\overline{a}|-\left(3\right)$

Comparing (2) & (3)

$\Rightarrow \frac{|\overline{c}{|}^2}{2}=|\overline{c}.\overline{a}|$

$\Rightarrow \frac{|\overline{c}{|}^2}{2}=|\overline{c}|.|\overline{a}|cos\theta$

($\because |(\overline{a}\times \overline{b}).\overline{c}|=\frac{|\overline{c}{|}^2}{2}$, $\therefore$ Maximum value of $|(\overline{a}\times \overline{b}).\overline{c}|$ is attained when $\overline{c}$ is maximum),

and for $|\overline{c}|$ to be maximum,

$cos\theta =1$

$\Rightarrow \theta =0$

$\therefore |\overline{c}|=2|\overline{a}|$

$\therefore |\overline{c}|=2$

$(\because |\overline{a}|=|\overline{b}|=1)$

Putting in (2)

$\Rightarrow |(\overline{a}\times \overline{b}).\overline{c}|=|(\overline{a}\times \overline{c}).\overline{b}|=\frac{|\overline{c}{|}^2}{2}=\frac{4}{2}$

$=2$

$\therefore$ B) answer