A) \[\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+\widehat{\mathbf{k}}\]
B) \[\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+\widehat{\mathbf{k}}\]
C) \[\frac{\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+\widehat{\mathbf{k}}}{\sqrt{3}}\]
D) \[\frac{\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}}}{\sqrt{3}}\]
Correct Answer: D
Solution :
Let\[\overset{\to }{\mathop{\mathbf{a}}}\,=3\widehat{\mathbf{i}}+\lambda \widehat{\mathbf{j}}+\widehat{\mathbf{k}}\]and\[\overset{\to }{\mathop{\mathbf{b}}}\,=2\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+8\widehat{\mathbf{k}}\] \[\overset{\to }{\mathop{\mathbf{b}}}\,=\widehat{\mathbf{i}}-\widehat{\mathbf{j}},\,\,\overset{\to }{\mathop{\mathbf{c}}}\,=\widehat{\mathbf{j}}+\widehat{\mathbf{k}}\] Since a unit vector is perpendicular to both the vectors. \[\therefore \] \[\widehat{\mathbf{a}}\cdot \overset{\to }{\mathop{\mathbf{b}}}\,=0\]and\[\widehat{\mathbf{a}}\cdot \overset{\to }{\mathop{\mathbf{c}}}\,=0\] \[\Rightarrow \] \[(x\widehat{\mathbf{i}}+y\widehat{\mathbf{j}}+z\widehat{\mathbf{k}})\cdot (\widehat{\mathbf{i}}-\widehat{\mathbf{j}})=0\] and \[(x\widehat{\mathbf{i}}+y\widehat{\mathbf{j}}+z\widehat{\mathbf{k}})\cdot (\widehat{\mathbf{j}}+\widehat{\mathbf{k}})=0\] \[\Rightarrow \] \[x-y=0\]and\[y+z=0\] \[\Rightarrow \] \[x=y\]and\[y=-z\] \[\Rightarrow \] \[x=y=-z=l\] (say) \[\therefore \] \[\widehat{\mathbf{a}}=\frac{l\widehat{\mathbf{i}}+l\widehat{\mathbf{j}}+l\widehat{\mathbf{k}}}{l\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}}=\frac{\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}}}{\sqrt{3}}\]
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