Question

Three vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are such that $\overrightarrow{a}\times \overrightarrow{b}=4(\overrightarrow{a}\times \overrightarrow{c})$ and $|\overrightarrow{a}|=|\overrightarrow{b}|=1$ and $|\overrightarrow{c}|=\frac{1}{4}$. If the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi }{3}$, then $\overrightarrow{b}$ is

• A. $\overrightarrow{a}+4\overrightarrow{c}$
• B. $\overrightarrow{a}-4\overrightarrow{c}$
• C. $4\overrightarrow{c}-\overrightarrow{a}$
• D. $2\overrightarrow{c}-\overrightarrow{a}$

Answer 1

Answer: (A), (C)

$4\overrightarrow{c}-\overrightarrow{a}$

$\overrightarrow{a}\times \overrightarrow{b}=4(\overrightarrow{a}\times \overrightarrow{c})$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=(\overrightarrow{a}\times 4\overrightarrow{c})$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}-(\overrightarrow{a}\times 4\overrightarrow{c})=0$
$\Rightarrow \overrightarrow{a}\times (\overrightarrow{b}-4\overrightarrow{c})=0$
So, $\overrightarrow{a}$ and $(\overrightarrow{b}-4\overrightarrow{c})$ are collinear or parallel vectors.
$\Rightarrow \lambda \overrightarrow{a}=(\overrightarrow{b}-4\overrightarrow{c})\dots \left[1\right]$
$\Rightarrow |\lambda \overrightarrow{a}{|}^2=|\overrightarrow{b}-4\overrightarrow{c}{|}^2$
$\Rightarrow {\lambda }^2|\overrightarrow{a}{|}^2=|\overrightarrow{b}{|}^2+16|\overrightarrow{c}{|}^2-8\times (\overrightarrow{b}\cdot \overrightarrow{c})$
$\Rightarrow \lambda =\pm 1$

Putting $\lambda =1$ in $\left[1\right]$, we get
$\overrightarrow{b}=4\overrightarrow{c}+\overrightarrow{a}$

Putting $\lambda =-1$ in $\left[1\right]$, we get
$\overrightarrow{b}=4\overrightarrow{c}-\overrightarrow{a}$