# Intersection of a Normal and a plane

• Jun 23rd 2009, 07:32 AM
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...
• Jun 23rd 2009, 07:42 AM
Plato
Quote:

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 $\displaystyle \left\{ \begin{gathered} x = 5 + 3t \hfill \\ y = 1 - t \hfill \\ z = 6 + 4t \hfill \\ \end{gathered} \right.$
Replace the $\displaystyle x,~y,~z~$ in the equation of the plane and solve for $\displaystyle t$.
• Jun 23rd 2009, 07:51 AM
dhiab
Quote:

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 )
• Jun 23rd 2009, 07:54 AM
dhiab
Quote:

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

Hello / you are z= 6+4t .... Thank you
• Jun 23rd 2009, 09:29 AM