Question #203626

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.


1
Expert's answer
2021-06-07T19:02:04-0400
x21=y+31=z3=t{x=t+2y=t3z=3t\frac{x-2}{1}=\frac{y+3}{-1}=\frac{z}{3}=t\\ \begin{cases} x=t+2\\ y=-t-3\\ z=3t \end{cases}


Substitute in the equation of the plane


2x3y+4z+4=02x-3y+4z+4=0



2t+4+3t+9+12t+4=02t+4+3t+9+12t+4=0

17t+17=0    t=117t+17=0 \implies t=-1\\

{x=1y=2z=3\begin{cases} x=1\\ y=-2\\ z=-3 \end{cases}

The distance of the point of intersection from the origin:


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

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