Question

If three vectors $a,b,c$ are such that $a\ne 0$ and $a\times b=2(a\times c),\left|a\right|=\left|c\right|=1,\left|b\right|=4$ and the angle between $b$ and $c$ is ${\cos }^{-1}\left(\frac{1}{4}\right)$. Also $b-2c=\lambda a$, then find the value of $\lambda$

• A. $\pm 4$
• B. $14$
• C. $\pm 2$
• D. $12$

Answer 1

The correct option is A $\pm 4$

Given, $a\times b=2(a\times c)$
$\Rightarrow a\times (b-2c)=0$

Thus $a$ is parallel to $b-2c$.
Now, $(b-2c)=\lambda a\Rightarrow {|b-2c|}^2={\lambda }^2{\left|a\right|}^2$
$\Rightarrow {\left|b\right|}^2+4{\left|c\right|}^2-4(b\cdot c)={\lambda }^2{\left|a\right|}^2$
$\Rightarrow 16+4-4\times \left|b\right|\left|c\right|\times \frac{1}{4}={\lambda }^2$
$\Rightarrow 20-4={\lambda }^2$

$\Rightarrow \lambda =\pm 4$