Question

Find the equation and length of perpendicular from a point `P(-3,0,-2)` to the line `(x-2)/2=(y-2)/(-2)=(z-1)/1`?

Answer 1

Equation is `(x+3)/3=y/4=(z+4)/2` and length of perpendicular is `sqrt29`

Explanation:

The equation of line is `(x-2)/2=(y-2)/(-2)=(z-1)/1=t` and its direction ratios are `2,-2,1` and a point on line is `Q(2t+2,-2t+2,t+1)`.

Direction ratios of point `P(-3,0,-2)` and `Q` are

`2t+2-(-3),-2t+2-0,t+1-(-2)` or `2t+5,-2t+2,t+3`

As it is perpendicular to the line, the dot product should be zero i.e.

`2(2t+5)-2(-2t+2)+1(t+3)=0`

or `4t+10+4t-4+t+3=0`

or `9t=-9` i.e. `t=-1`

and coordinates of point `Q_0` are `(0,4,0)`

length of `PQ_0` is `sqrt((0-(-3))^2+(4-0)^2+(0-(-2))^2)`

= `sqrt(9+16+4)=sqrt29`

Equation of a line joining two points `(x_1,y_1,z_1)` and `(x_2,y_2,z_2)` is `(x-x_1)/(x_2-x_1)=(y-y_1)/(y_2-y_1)=(z-z_1)/(z_2-z_1)`

Equation of `PQ_0` is

`(x-(-3))/(0-(-3))=(y-0)/(4-0)=(z-(-2))/(0-(-2))`

or `(x+3)/3=y/4=(z+4)/2`