A) \[\frac{2\hat{j}+\hat{k}}{\sqrt{5}}\]
B) \[\frac{3\hat{i}+2\hat{j}-2\hat{k}}{\sqrt{17}}\]
C) \[\frac{3\hat{i}+2\hat{j}+2\hat{k}}{\sqrt{17}}\]
D) \[\frac{2\hat{i}+2\hat{j}-2\hat{k}}{3}\]
Correct Answer: D
Solution :
Let \[x\hat{i}+y\hat{j}+z\hat{k}\] be the required unit vector. Since \[\hat{a}\] is perpendicular to \[(2\hat{i}-\hat{j}+2\hat{k}).\] \[\therefore \]\[2x-y+2z=0\] ..... (i) Since vector \[x\hat{i}+y\hat{j}+z\hat{k}\] is coplanar with the vector \[\hat{i}+\hat{j}-\hat{k}\] and \[2\hat{i}+2\hat{j}-\hat{k}\] \[\therefore \]\[x\hat{i}+y\hat{j}+z\hat{k}\] \[=p(\hat{i}+\hat{j}-\hat{k})+q(2\hat{i}+2\hat{j}-\hat{k}),\]where p and q are some scalars. \[\Rightarrow \]\[x\hat{i}+y\hat{j}+z\hat{k}\] \[=(p+2q)\hat{i}+(p+2q)\hat{j}-(p+q)\hat{k}\] \[\Rightarrow \]\[x=p+2q,y=p+2q,z=-p-q\] Now from equation (i), \[2p+4q-p-2q-2p-2q=0\] \[\Rightarrow \]\[-p=0\Rightarrow p=0\] \[\therefore \]\[x=2q,y=2q,z=-q\] Since vector \[x\hat{i}+y\hat{j}+z\hat{k}\] is a unit vector, therefore\[|x\hat{i}+y\hat{j}+z\hat{k}|=1\] \[\Rightarrow \]\[\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=1\]\[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1\] \[\Rightarrow \]\[4{{q}^{2}}+4{{q}^{2}}+{{q}^{2}}=1\]\[\Rightarrow \]\[9{{q}^{2}}=1\Rightarrow q=\pm \frac{1}{3}\]\[\Rightarrow \] When\[q=\frac{1}{3},\]then\[x=\frac{2}{3},y=\frac{2}{3},z=-\frac{1}{3}\] When \[q=-\frac{1}{3},\]then \[x=-\frac{2}{3},y=-\frac{2}{3},z=\frac{1}{3}\] Here required unit vector is \[\frac{2}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\] or\[-\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}.\]
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