Question

If for vectors A and B, $\mathrm{A}.\mathrm{B}=\left|\mathrm{A}\times \mathrm{B}\right|$, then $\left|\mathrm{A}-\mathrm{B}\right|$ is

• A. $\sqrt{\left({\mathrm{A}}^2+{\mathrm{B}}^2\sqrt{(2\mathrm{AB}})\right)}$
• B. $\sqrt{\left({\mathrm{A}}^2+{\mathrm{B}}^2-\sqrt{\left(2\mathrm{AB}\right)}\right)}$
• C. $\sqrt{\left({\mathrm{A}}^2+{\mathrm{B}}^2-\sqrt{2}\mathrm{AB}\right)}$
• D. $\sqrt{\left({\mathrm{A}}^2+{\mathrm{B}}^2+\sqrt{2}\mathrm{AB}\right)}$

Answer 1

The correct option is C

$\sqrt{\left({\mathrm{A}}^2+{\mathrm{B}}^2-\sqrt{2}\mathrm{AB}\right)}$

Explanation for the correct answer:

Given, $\mathrm{A}.\mathrm{B}=\left|\mathrm{A}\times \mathrm{B}\right|$

$\begin{array}{rcl}& \Rightarrow & \cos \mathrm{\theta }=\sin \mathrm{\theta }\\ & \Rightarrow & \tan \mathrm{\theta }=1\\ & \Rightarrow & \mathrm{\theta }=\frac{\mathrm{\pi }}{4}\end{array}$

$\begin{array}{rcl}{\left|\mathrm{A}-\mathrm{B}\right|}^2& =& {\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{AB}\\ & =& {\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{AB}\cos \left(\frac{\mathrm{\pi }}{4}\right)\\ & =& {\mathrm{A}}^2+{\mathrm{B}}^2-\sqrt{2}\mathrm{AB}\\ & \Rightarrow & \left|\mathrm{A}-\mathrm{B}\right|=\sqrt{({\mathrm{A}}^2+{\mathrm{B}}^2-\sqrt{2}\mathrm{AB})}\end{array}$

Hence, the correct answer is option (C).