Let $\overrightarrow{\mathbf{a}}\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{,}\overrightarrow{\mathbf{c}}$ be the vectors such that$\ \overrightarrow{\mathbf{a}}
eq 0$ and $\overrightarrow{\mathbf{a}}\mathbf{ imes}\overrightarrow{\mathbf{b}} = 2\overrightarrow{\mathbf{a}}\mathbf{ imes}\overrightarrow{\mathbf{c}}\mathbf{,}\left| \overrightarrow{\mathbf{a}} ight| = \left| \overrightarrow{\mathbf{c}} ight| = 1,\left| \overrightarrow{\mathbf{b}} ight| = 4$ and $\left| \overrightarrow{\mathbf{b}}\mathbf{ imes}\overrightarrow{\mathbf{c}} ight| = \sqrt{15}$. If $\overrightarrow{\mathbf{b}}\mathbf{-}2\overrightarrow{\mathbf{c}}\mathbf{=}\lambda\mathbf{\ }\overrightarrow{\mathbf{a}}$, then $\lambda$ is equal to

Question

Let $\overrightarrow{\mathbf{a}}\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{,}\overrightarrow{\mathbf{c}}$ be the vectors such that$\ \overrightarrow{\mathbf{a}} 
eq 0$ and $\overrightarrow{\mathbf{a}}\mathbf{	imes}\overrightarrow{\mathbf{b}} = 2\overrightarrow{\mathbf{a}}\mathbf{	imes}\overrightarrow{\mathbf{c}}\mathbf{,}\left| \overrightarrow{\mathbf{a}} ight| = \left| \overrightarrow{\mathbf{c}} ight| = 1,\left| \overrightarrow{\mathbf{b}} ight| = 4$ and $\left| \overrightarrow{\mathbf{b}}\mathbf{	imes}\overrightarrow{\mathbf{c}} ight| = \sqrt{15}$. If $\overrightarrow{\mathbf{b}}\mathbf{-}2\overrightarrow{\mathbf{c}}\mathbf{=}\lambda\mathbf{\ }\overrightarrow{\mathbf{a}}$, then $\lambda$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given that, $\left| \overrightarrow{\mathbf{a}} ight| = \left| \overrightarrow{\mathbf{c}} ight| = 1,\left| \overrightarrow{\mathbf{b}} ight| = 4$
Let angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\alpha$, then
$\left| \overrightarrow{\mathbf{b}}\mathbf{	imes}\overrightarrow{\mathbf{c}} ight| = \sqrt{15}$ (given)
$\Rightarrow \ \left| \overrightarrow{\mathbf{b}} ight|\left| \overrightarrow{\mathbf{c}} ight|\sin{\alpha =}\sqrt{15}$
$\Rightarrow \sin{\alpha = \frac{\sqrt{15}}{4 	imes 1} = \frac{\sqrt{15}}{4}}$
$	herefore\cos{\alpha = \sqrt{1 - \sin^{2}\alpha} = \frac{1}{4}}$
We have, $\overrightarrow{\mathbf{b}} = 2\overrightarrow{\mathbf{c}} = \lambda\ \overrightarrow{\mathbf{a}}$
On squaring both sides, we get
$\left( \ \overrightarrow{\mathbf{b}} - 2\overrightarrow{\mathbf{c}} ight)^{2} = \lambda^{2}\left( \overrightarrow{\mathbf{a}} ight)^{2}$
$\Rightarrow {\overrightarrow{\mathbf{b}}}^{2} + 4{\overrightarrow{\mathbf{c}}}^{2} - 4\ \overrightarrow{\mathbf{b}}\mathbf{\bullet}\overrightarrow{\mathbf{c}} = \lambda^{2}{\ \overrightarrow{\mathbf{a}}}^{2}$
$\Rightarrow 16 + 4 - 4\left| \overrightarrow{\mathbf{b}} ight|\left| \overrightarrow{\mathbf{c}} ight|\cos{\alpha =}\lambda^{2}$
$\Rightarrow \ 16 + 4 - 4 	imes 4 	imes 1 	imes \frac{1}{4} = \lambda^{2}$
$\Rightarrow \lambda^{2} = 16 + 4 - 4 = 16\$
$\Rightarrow \ \lambda = \pm 4$

Explanation

Reviewed by Dr. Abubacker

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