Question

Let $a,b$ and $c$ be three vectors satisfying $a\times b=(a\times c)$, $\left|a\right|=\left|c\right|=1,\left|b\right|=4$ and $|b\times c|=\sqrt{15}$. If $b-2c=\lambda a$, then $\lambda$ equals

• A. $1$
• B. $-1$
• C. $2$
• D. $-4$

Answer 1

The correct option is D $-4$

Let $\theta$ be the angle between $b$ and $c$
given, $|b\times c|=\sqrt{15}$

$\Rightarrow \left|b\right|\left|c\right|\sin \theta =\sqrt{15}$

$\Rightarrow \sin \theta =\frac{\sqrt{15}}{\left|b\right|\left|c\right|}$

$\Rightarrow \sin \theta =\frac{\sqrt{15}}{4\times 1}=\frac{\sqrt{15}}{4}$
Therefore, $\cos \theta =\sqrt{1-\frac{15}{16}}=\frac{1}{4}$
Now given, $b-2c=\lambda a\Rightarrow |b-2c^2|=\left|\lambda a^2\right|$
$b-2c=\lambda a\Rightarrow |b-2c^2|=\left|\lambda a^2\right|$

$\Rightarrow 16+4-4\left|b\right|\left|c\right|\cos \theta ={\lambda }^2$

$\Rightarrow {\lambda }^2=16(\cos \theta =\frac{1}{4})$

$\Rightarrow \lambda =\pm 4$