Question

please solve ASAP:(

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

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

Substituting the values of x, y and z in the equation of the plane

$$2x-3y+4z+4=0$$

We get,

$$2t+4+3t+9+12t+4=0$$ $$\therefore17t+17=0 \implies t=-1\\$$ $$\begin{cases} x=1\\ y=-2\\ z=-3 \end{cases}$$

Hence, the distance of the point of intersection of line and plane from the origin is

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