A) \[\mathbf{a}=2\,\mathbf{i}+2\,\mathbf{j}+3\,\mathbf{k},\,\,\mathbf{b}=-\mathbf{i}+2\,\mathbf{j}+\mathbf{k}\]
B) \[\frac{3}{4}\]
C) \[\frac{3}{2}\]
D) \[\frac{4}{3}\]
Correct Answer: B
Solution :
Since \[\mathbf{a}+\lambda \mathbf{b}\]is perpendicular to \[\mathbf{a}-\lambda \mathbf{b}\], then their product will be zero. So, \[(\mathbf{a}+\lambda \mathbf{b}).(\mathbf{a}-\lambda \mathbf{b})=0\] Þ \[|\mathbf{a}{{|}^{2}}-{{\lambda }^{2}}|\mathbf{b}{{|}^{2}}=0\] or \[{{\lambda }^{2}}=\frac{|\mathbf{a}{{|}^{2}}}{|\mathbf{b}{{|}^{2}}}\Rightarrow {{\lambda }^{2}}=\frac{9}{16}\] or \[\lambda =\pm \frac{3}{4}\], \[[\because \,|\mathbf{a}|=3,|\mathbf{b}|=4]\].
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