Question

If $\overset{ \rightarrow }{m}_{a}, \, \overset{ \rightarrow }{m}_{b}$ and $\overset{ \rightarrow }{m}_{c}$ are $3$ unit vectors such that $\overset{ \rightarrow }{m}_{a}\cdot \overset{ \rightarrow }{m}_{b}=0, \, \overset{ \rightarrow }{m}_{a}\cdot \overset{ \rightarrow }{m}_{c}=0$ and the angle between $\overset{ \rightarrow }{m}_{b}$ and $\overset{ \rightarrow }{m}_{c}$ is $\frac{\pi }{3},$ then the value of $\left|\overset{ \rightarrow }{m}_{a} \times \overset{ \rightarrow }{m}_{b} - \overset{ \rightarrow }{m}_{a} \times \overset{ \rightarrow }{m}_{c}\right|$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$\because \vec{m}_{a} \cdot \vec{m}_{b}=0, \vec{m}_{a} \cdot \vec{m}_{c}=0$
$\Rightarrow \vec{m}_{a} \cdot\left(\vec{m}_{b}-\vec{m}_{c}\right)=0 \Rightarrow \vec{m}_{a}$ is perpendicular to $\vec{m}_{b}-\vec{m}_{c}$
Therefore, $\left|\vec{m}_{a} \times\left(\vec{m}_{b}-\vec{m}_{c}\right)\right|=\left|\vec{m}_{a} \| \vec{m}_{b}-\vec{m}_{c}\right|$ $=1 \times \sqrt{1+1-2 \times 1 \times 1 \times \cos \frac{\pi}{3}}$ $=1$