Question

If ${\overrightarrow{m}}_a,{\overrightarrow{m}}_b$ and ${\overrightarrow{m}}_c$ are $3$ unit vectors such that ${\overrightarrow{m}}_a\cdot {\overrightarrow{m}}_b=0,{\overrightarrow{m}}_a\cdot {\overrightarrow{m}}_c=0$ and the angle between ${\overrightarrow{m}}_b$ and ${\overrightarrow{m}}_c$ is $\frac{\pi }{3},$ then the value of $\left|{\overrightarrow{m}}_a\times {\overrightarrow{m}}_b-{\overrightarrow{m}}_a\times {\overrightarrow{m}}_c\right|$ 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, $\left|{\vec{m}}_a\times \left({\vec{m}}_b-{\vec{m}}_c\right)\right|=\left|{\vec{m}}_a\Vert {\vec{m}}_b-{\vec{m}}_c\right|$ $=1\times \sqrt{1+1-2\times 1\times 1\times \cos \frac{\pi }{3}}$ $=1$