A) -2 only
B) \[\pm 2\]
C) 3 only
D) \[\pm 3\]
Correct Answer: B
Solution :
[b] As given: \[\overset{\to }{\mathop{a}}\,=\hat{i}+2\hat{j}-3\hat{k}\] and \[\overset{\to }{\mathop{b}}\,=3\hat{i}-\hat{j}+\lambda \hat{k}\] \[\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,=\hat{i}+2\hat{j}-3\hat{k}+3\hat{i}-\hat{j}+\lambda \hat{k}\] \[=4\hat{i}+\hat{j}+(\lambda -3)\hat{k}\] and \[\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{b}}\,=\hat{i}+2\hat{j}-3\hat{k}-3\hat{i}+\hat{j}-\lambda \hat{k}\] \[=-2\hat{i}+3\hat{j}-(3+\lambda )\hat{k}\] \[(\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,)\] is perpendicular to \[(\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{b}}\,)\] \[\Rightarrow (\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,).(\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{b}}\,)=0\] \[\Rightarrow \{4\hat{i}+\hat{j}+(\lambda -3)\hat{k}\}\{-2\hat{i}+3\hat{j}-(3-\lambda )\hat{k}\}=0\] \[\Rightarrow -8+3+({{3}^{2}}-{{\lambda }^{2}})=0\] \[\Rightarrow -4-{{\lambda }^{2}}=0\] \[\Rightarrow \lambda =\pm \,2\]
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