CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image