A) \[\frac{\mathbf{i}}{2}+\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}}\]
B) \[\frac{\mathbf{i}}{2}+\frac{\mathbf{j}}{2}-\frac{\mathbf{k}}{\sqrt{2}}\]
C) \[-\frac{\mathbf{i}}{2}-\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}}\]
D) None of these
Correct Answer: C
Solution :
Let \[\mathbf{a}=l\mathbf{i}+m\mathbf{j}+n\mathbf{k},\] where \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1.\] a makes an angle \[\frac{\pi }{4}\] with \[z-\]axis. \[\therefore \,\,n=\frac{1}{\sqrt{2}},\] \[{{l}^{2}}+{{m}^{2}}=\frac{1}{2}\] ?..(i) \[\therefore \,\,\mathbf{a}=l\,\mathbf{i}+m\,\mathbf{j}+\frac{\mathbf{k}}{\sqrt{2}}\] \[\mathbf{a}+\mathbf{i}+\mathbf{j}=(l+1)\mathbf{i}+(m+1)\mathbf{j}+\frac{\mathbf{k}}{\sqrt{2}}\] Its magnitude is 1, hence \[{{(l+1)}^{2}}+{{(m+1)}^{2}}=\frac{1}{2}\] .....(ii) From (i) and (ii), \[2lm=\frac{1}{2}\Rightarrow l=m=-\frac{1}{2}\] Hence \[\mathbf{a}=-\frac{\mathbf{i}}{2}-\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}}\].
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