A) \[0{}^\circ \]
B) \[\pi \]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
The scalar product of two vectors is \[\overrightarrow{\mathbf{A}}\mathbf{.}\overrightarrow{\mathbf{B}}\mathbf{=}AB\cos \theta \] Given, \[\mathbf{|}\overrightarrow{\mathbf{P}}\mathbf{+}\overrightarrow{\mathbf{Q}}\mathbf{|=|}\overrightarrow{\mathbf{P}}\mathbf{-}\overrightarrow{\mathbf{Q}}\mathbf{|}\] Squaring both sides of the equation, we get \[\mathbf{|}\overrightarrow{\mathbf{P}}\mathbf{+}\overrightarrow{\mathbf{Q}}{{\mathbf{|}}^{2}}\mathbf{=|}\overrightarrow{\mathbf{P}}\mathbf{-}\overrightarrow{\mathbf{Q}}{{\mathbf{|}}^{2}}\] \[\overrightarrow{{{\mathbf{P}}^{\mathbf{2}}}}\mathbf{+}\overrightarrow{{{\mathbf{Q}}^{\mathbf{2}}}}\mathbf{+}\overrightarrow{{{\mathbf{P}}^{\mathbf{2}}}}\mathbf{+}\overrightarrow{{{\mathbf{Q}}^{\mathbf{2}}}}\mathbf{-2}\overrightarrow{\mathbf{P}}\mathbf{.}\overrightarrow{\mathbf{Q}}\cos \theta \] \[\Rightarrow \] \[4\overrightarrow{\mathbf{P}}.\overrightarrow{\mathbf{Q}}=0\] \[\Rightarrow \] \[4\overrightarrow{\mathbf{P}}.\overrightarrow{\mathbf{Q}}\cos \theta =0\] \[\Rightarrow \] \[\cos \theta =0\] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\] Hence, the two vectors are perpendicular to each other.
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