Question

# From the point $$P(3, -1, 11)$$, a perpendicular is drawn on the line $$L$$ given by the equation $$\dfrac {x}{2} = \dfrac {y - 2}{3} = \dfrac {z - 3}{4}$$. Let $$Q$$ be the foot of the perpendicular.What are the direction ratios of the line segment $$PQ$$?

A
(1,6,4)
B
(1,6,4)
C
(1,6,4)
D
(2,6,4)

Solution

## The correct option is B $$(-1, 6, -4)$$The general point on the line $$\dfrac { x }{ 2 } =\dfrac { y-2 }{ 3 } =\dfrac { z-3 }{ 4 }$$ is $$(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$$Let $$Q(x,y,z)=(2\lambda ,3\lambda +2,4\lambda +3)$$The directional cosines of $$PQ= 2\lambda -3,3\lambda +3,4\lambda -8$$Since the line $$L$$ and $$PQ$$ are perpendicular$$(2\lambda -3)(2)+(3\lambda +3)(3)+(4\lambda -8)4=0\\ \Longrightarrow \lambda =1\\ \Longrightarrow Q=(2,5,7)$$The direction ratios of $$PQ=(-1,6,-4)$$Mathematics

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