Answer to Question #91546 in Analytic Geometry for Ra

Question

Find the distance of the point of intersection of the line
x-2/1=y+3/-1=z/3
and the plane 2x-3y+4z+4=0 from the origin.

Answer 1

$$\frac{x-2}{1}=\frac{y+3}{-1}=\frac{z}{3}=t\\ \begin{cases} x=t+2\\ y=-t-3\\ z=3t \end{cases}$$

then

$$2x-3y+4z+4=\\ =2t+4+3t+9+12t+4=17t+17=0\\ \implies t=-1\\ \begin{cases} x=1\\ y=-2\\ z=-3 \end{cases}$$

and the distance from the origin:

$$r=\sqrt{(1)^2+(-2)^2+(-3)^2} = \sqrt{14}$$