A) \[-\frac{1}{\sqrt{2}}\widehat{\mathbf{i}}+\frac{1}{\sqrt{2}}\widehat{\mathbf{k}}\]
B) \[\frac{1}{\sqrt{2}}\mathbf{\hat{i}}-\frac{1}{\sqrt{2}}\mathbf{\hat{k}}\]
C) \[\frac{1}{3\sqrt{2}}\widehat{\mathbf{i}}+\frac{4}{3\sqrt{2}}\widehat{\mathbf{j}}+\frac{1}{3\sqrt{2}}\widehat{\mathbf{k}}\]
D) None of the above
Correct Answer: B
Solution :
Let the required vector be\[\overset{\to }{\mathop{\mathbf{c}}}\,=x\widehat{\mathbf{i}}+y\widehat{\mathbf{k}}\] As it is a unit vector, therefore \[|\overset{\to }{\mathop{\mathbf{c}}}\,|\,\,=1\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}=1\] ? (i) \[\overset{\to }{\mathop{\mathbf{a}}}\,\]and\[\overset{\to }{\mathop{\mathbf{c}}}\,\]are inclined at the angle\[{{45}^{o}}\]. \[\therefore \] \[\cos {{45}^{o}}=\frac{2x-y}{\sqrt{4+4+1}}\] \[\Rightarrow \] \[2x-y=\frac{3}{\sqrt{2}}\] ? (ii) \[\overset{\to }{\mathop{\mathbf{b}}}\,\]and\[\overset{\to }{\mathop{\mathbf{c}}}\,\]are inclined at an angle\[{{60}^{o}}\]. \[\therefore \] \[-\frac{y}{\sqrt{2}}=\cos {{60}^{o}}\Rightarrow \,\,y=-\frac{1}{\sqrt{2}}\] ? (iii) From Eqs. (ii) and (iii), we get\[x=\frac{1}{\sqrt{2}}\] Hence, the required vector is\[\frac{1}{\sqrt{2}}\widehat{\mathbf{i}}-\frac{1}{\sqrt{2}}\widehat{\mathbf{k}}\].
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