A) Acute
B) Obtuse
C) \[\frac{\pi }{2}\]
D) \[\pi \]
Correct Answer: A
Solution :
\[|\mathbf{a}+\mathbf{b}|\,>\,|\mathbf{a}-\mathbf{b}|\] Squaring both sides, we get \[{{a}^{2}}+{{b}^{2}}+2\mathbf{a}\,.\,\mathbf{b}\,>\,{{a}^{2}}+{{b}^{2}}-2\mathbf{a}\,.\,\mathbf{b}\] \[\] Þ \[4\mathbf{a}.\mathbf{b}>0\] Þ \[\cos \theta >0\]. Hence \[\theta \,<\,90{}^\circ \], (acute).
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