A) \[{{0}^{o}}\]
B) \[\pi \]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
The scalar product of two vector's is \[\vec{A}.\vec{B}=AB\,\,\cos \theta \] Given, \[|\vec{P}+\vec{Q}|=|\vec{P}-\vec{Q}|\] Squaring both sides of the equation, we get \[|\vec{P}+\vec{Q}{{|}^{2}}=|\vec{P}-\vec{Q}{{|}^{2}}\] \[{{\vec{P}}^{2}}+{{\vec{Q}}^{2}}+2\vec{P}.\vec{Q}={{\vec{P}}^{2}}+{{\vec{Q}}^{2}}-2\vec{P}.\vec{Q}\,\cos \theta \] \[\Rightarrow \] \[4\vec{P}.\vec{Q}=0\] \[\Rightarrow \] \[4\vec{P}.\vec{Q}\,\cos \theta =0\] \[\Rightarrow \] \[\cos \theta =0\] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\] Hence, the two vectors are perpendicular to each other.
You need to login to perform this action.
You will be redirected in
3 sec
