A) \[\pi /2\]
B) \[\pi /4\]
C) \[\pi /3\]
D) 0
Correct Answer: A
Solution :
Let \[\mathbf{a}=2\mathbf{i}+3\mathbf{j}+\mathbf{k}\] and \[\mathbf{b}=2\mathbf{i}-\mathbf{j}+\mathbf{k}\] Since \[\cos \,\theta =\frac{\mathbf{a}\,.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\] \[=\frac{(2\mathbf{i}+3\mathbf{j}+\mathbf{k})\,.\,(2\mathbf{i}-\mathbf{j}-\mathbf{k})\,}{\sqrt{{{(2)}^{2}}+{{(3)}^{2}}+{{(1)}^{2}}}\sqrt{{{(2)}^{2}}+{{(-1)}^{2}}+{{(-1)}^{2}}}}\] \[=\frac{4-3-1}{\sqrt{(4+9+1)}\sqrt{(4+1+1)}}=0\] \[\therefore \theta =\frac{\pi }{2}\].
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