# Thread: Intersection of a plane and a normal.

1. ## Intersection of a plane and a normal.

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

I haven't solved a question like this before in my book, so I am not too sure on how to go about this question. I know that the normal vector to $\displaystyle 3x-y+4z=12$ is $\displaystyle \vec{n} = (3, -1, 4)$. So, to find the point of intersection, do I take the cross product of the normal and the given point? Or, would I have to find another point on the plane (I'll let $\displaystyle x = 2$ and $\displaystyle y = 1$ to get
$\displaystyle 3(2) - (1)+4z = 12 => 4z = 7 => z = \frac{7}{4}$, this point will be $\displaystyle B(2, 1,\frac{7}{4})$ and then find $\displaystyle \vec{AB}$ and take the cross product of $\displaystyle \vec{n}$ and $\displaystyle \vec{AB}$?

I just never encountered a question like this before, so I would appreciate some help on getting started. Thanks in advance.

2. Find where the line
$\displaystyle \ell (t) = \left\{ {\begin{array}{*{20}c} {x = 5 + 3t} \\ {y = 1 - t} \\ {z = 6 + 4t} \\ \end{array} } \right.$
intersects the plane.

3. Originally Posted by Plato
Find where the line
$\displaystyle \ell (t) = \left\{ {\begin{array}{*{20}c} {x = 5 + 3t} \\ {y = 1 - t} \\ {z = 6 + 4t} \\ \end{array} } \right.$
intersects the plane.
So, substituting the normal $\displaystyle (3, -1, 4)$:
$\displaystyle 3 = 5 + 3t => -2 = 3t => x = -\frac{2}{3}$
$\displaystyle -1 = 1 - t => y = -2$
$\displaystyle 4 = 6 + 4t => -2 = 4t => z = -\frac{1}{2}$

Therefore, the line intersects the plane at the point $\displaystyle (-\frac{2}{3}, -2, -\frac{1}{2}$, correct?

4. No, that's not the point. The normal is a vector, not an (x, y, z) point in the plane. Also note that you have made no use at all of the "= 12" part of the equation of the plane.

Instead, put x= 5+ 3t, y= 1- t, and z= 6+ 4t in for the x, y, and z in the equation of the plane. Then solve for t and use that to find x, y, and z.

5. Ah, I see. I'll reattempt this:
$\displaystyle 3(5 + 3t) - (1-t) + 4 (6 + 4t) = 12$
$\displaystyle 15 + 9t - 1 + t +24 + 16t = 12$
$\displaystyle 26t = -26$
$\displaystyle t = -1$
Substituting this back into the original parametric equation:
$\displaystyle x = 5 + 3(-1) = 2$
$\displaystyle y = 1 - (-1) = 2$
$\displaystyle z = 6 + 4(-1) = 2$
Point of intersection of the normal and the plane is (2, 2, 2). Correct?

6. Check it yourself. Does x=2, y= 2, z= 2 satisfy the equation of the plane, 3x- y+ 4z= 12 (so that (2, 2, 2) is on theplane)? Does there exist a single value of t such that x= 5+ 3t= 2, y= 1- t= 2, and z= 6+ 4t= 2 (so that (2, 2, 2) is on the normal line)?

7. Originally Posted by HallsofIvy
Check it yourself. Does x=2, y= 2, z= 2 satisfy the equation of the plane, 3x- y+ 4z= 12 (so that (2, 2, 2) is on theplane)? Does there exist a single value of t such that x= 5+ 3t= 2, y= 1- t= 2, and z= 6+ 4t= 2 (so that (2, 2, 2) is on the normal line)?
Thank you for all the help and effort, HallsofIvy. I now understand perfectly.