A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: A
Solution :
Projection of \[\vec{A}=3\hat{i}-\hat{j},\] Projection of \[\vec{B}=-3\hat{i}+5\hat{j}+\sqrt{2}\hat{k}\] \[\therefore \] \[\overrightarrow{AB}=-6\hat{i}+6\hat{j}+\sqrt{2}\hat{k}\] Let \[\overrightarrow{CD}=\hat{i}/2+\hat{j}/2+\hat{k}/\sqrt{2}\] Now, Projection of \[\overrightarrow{AB}\] on \[\overrightarrow{CD}=\frac{\overrightarrow{AB}.\overrightarrow{CD}}{|\overrightarrow{CD}|}\] \[=\frac{(-6\hat{i}+6\hat{i}+\sqrt{2}\hat{k}).\,\left( \frac{{\hat{i}}}{2}+\frac{{\hat{j}}}{2}+\frac{{\hat{k}}}{\sqrt{2}} \right)}{\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{2}}}\] \[=\frac{-3+3+1}{\sqrt{\frac{4}{4}}}=1\]
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