Question

If three vectors $a,b,c$ are such that $a\ne 0$ and $a\times b=2(a\times c),|a|=|c|=1,|b|=4$ 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

Answer: (A)

$a\times b=2(a\times c)$
$\Rightarrow a\times (b-2c)=0$
$\Rightarrow a$ is parallel to $b-2c$
Now, $(b-2c)=\lambda a$
$\Rightarrow |b-2c{|}^2={\lambda }^2|a{|}^2$
$\Rightarrow |b{|}^2+4|c{|}^2-4(b\cdot c)={\lambda }^2|a{|}^2$
$\Rightarrow 16+4-4\times |b||c|\times \frac{1}{4}={\lambda }^2$
$\Rightarrow 20-4={\lambda }^2$
$\Rightarrow \lambda =\pm 4$