# Math Help - Intersection of a Normal and a plane

1. ## Intersection of a Normal and a plane

A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.

So from what I know the normal would be:
n = (3,-1,4)

But I don't know how to go about answering this....Any help is appreciated...

A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.
The normal line through A is $\left\{ \begin{gathered} x = 5 + 3t \hfill \\ y = 1 - t \hfill \\ z = 6 + 4t \hfill \\ \end{gathered} \right.$
Replace the $x,~y,~z~$ in the equation of the plane and solve for $t$.

A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.

So from what I know the normal would be:
n = (3,-1,4)

But I don't know how to go about answering this....Any help is appreciated...
Hello : the nomal is definid by : x-5=3t
y-1=-t
z-6=4t
x=5+3t and y=1-t and z=6+4t
Remplace in plan equation : 3(5+3t) -(1-t)+4(6+4t)=12
solve : I'have t= -1
remplace in system : x=5+3( -1 ) and y =1-(-1 ) and z= 6+4( -1)
coclusion: the point M( 2 , 2 , 2 )

4. Originally Posted by Plato
The normal line through A is $\left\{ \begin{gathered} x = 5 + 3t \hfill \\ y = 1 - t \hfill \\ z = 6 + 5t \hfill \\ \end{gathered} \right.$
Replace the $x,~y,~z~$ in the equation of the plane and solve for $t$.
Hello / you are z= 6+4t .... Thank you

5. I understand now, thank you so much