A) \[{{0}^{{}^\circ }}\]
B) \[{{45}^{{}^\circ }}\]
C) \[{{60}^{{}^\circ }}\]
D) \[{{90}^{{}^\circ }}\]
Correct Answer: D
Solution :
[d] \[\because \left| \overrightarrow{a}+\overrightarrow{b} \right|=\left| \overrightarrow{a}-\overrightarrow{b} \right|\] Squaring both sides, we have \[{{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+2\left| \overrightarrow{a} \right|.\left| \overrightarrow{b} \right|.\cos \theta {{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}-2\left| \overrightarrow{a} \right|.\left| \overrightarrow{b} \right|\cos \theta \]\[\Rightarrow 4ab.cos\theta =0~~\Rightarrow cos\theta =cos\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{2}\] Hence, option [d] is correct.
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