A) \[2\overrightarrow{OA}+3\overrightarrow{OB}\]
B) \[4\overrightarrow{OC}\]
C) \[-\overrightarrow{OC}\]
D) \[\overrightarrow{OC}\]
Correct Answer: A
Solution :
\[\overrightarrow{c}\]coplanar with\[\overrightarrow{a},\overrightarrow{b}\] \[\therefore \] \[\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}\] \[\Rightarrow \] \[\overrightarrow{c}=x(2\hat{i}+\hat{j}+\hat{k})+y(\hat{i}+2\hat{j}-\hat{k})\] \[\overrightarrow{c}=(2x+y)\hat{i}+(x+2y)\hat{j}+(x-y)\hat{k}\] \[\because \] \[\overrightarrow{a}.\overrightarrow{b}=0\] \[\Rightarrow \] \[2(2x+y)+x+2y+x-y=0\] \[\Rightarrow \] \[y=-2x\] \[\therefore \] \[\overrightarrow{c}=-3x\hat{j}+3x\hat{k}=3x(-\hat{j}+\hat{k})\] \[\because \] \[|c|=1\] \[\Rightarrow \] \[9{{x}^{2}}+9{{x}^{2}}=1\] \[\Rightarrow \] \[x=\pm \frac{1}{3\sqrt{2}}\] \[\Rightarrow \] \[\overrightarrow{c}=\frac{1}{\sqrt{2}}(-\hat{j}+\hat{k})\]You need to login to perform this action.
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