A perpendicular is drawn from a point on the line x-12=y+1-1= z1 to the plane x+y+z=3 such…

Question

A perpendicular is drawn from a point on the line $\dfrac{x-1}2=\dfrac{y+1}{-1}=\dfrac z1$ to the plane $x+y+z=3$ such that the foot of the perpendicular $Q$ also lies on the plane $x-y+z=3$. Then the co-ordinates of $Q$ are:

Accepted Answer

$${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1} = \lambda $$

Let a point P on the line is

(2$$\lambda $$ + 1, -- $$\lambda $$ --1, + $$\lambda $$)

Foot of $${ \bot ^r}Q$$ is given by

$${{x - 2\lambda  - 1} \over 1} = {{y + \lambda  + 1} \over 1} = {{z - \lambda } \over 1} =  - {{\left( {2\lambda  - 3} \right)} \over 3}$$

$$ \therefore $$  Q lies on x + y + z = 3 \& x -- y + z = 3

$$ \Rightarrow $$  x + z = 3 \& y = 0

$$ \therefore $$ $$y = 0 \Rightarrow \lambda  + 1 = {{ - 2\lambda  + 3} \over 3} \Rightarrow \lambda  = 0$$

$$ \therefore $$ Q is (2, 0, 1)

Solutions · 1