question_answer

Projection of line segment joining (2, 3, 4) and (5, 6, 7) on plane \[2x+y+z=1\] is

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,             \[AC=\sqrt{3\sqrt{3}{{)}^{2}}-{{(2\sqrt{6})}^{2}}}=\sqrt{3}\]