A) 2
B) \[\sqrt{3}\]
C) 3
D) \[3\sqrt{3}\]
Correct Answer: B
Solution :
[b] Let \[A\equiv (2,3,4)\] and \[B\equiv (5,6,7)\] Then \[AB=3\sqrt{3}.\] Plane: \[2x+y+z=1\] Projection of \[\overrightarrow{AB}\] on vector \[2\hat{i}+\hat{j}+\hat{k}\] which is normal to plane, \[BC=\frac{(3\hat{i}+3\hat{j}+3\hat{k}).(2\hat{i}+\hat{j}+\hat{k})}{\sqrt{6}}=\frac{12}{\sqrt{6}}=2\sqrt{6}\] \[\therefore \]Projection of \[\overrightarrow{AB}\]on plane, \[AC=\sqrt{3\sqrt{3}{{)}^{2}}-{{(2\sqrt{6})}^{2}}}=\sqrt{3}\]You need to login to perform this action.
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