Intersection of a plane and a normal.

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June 4th 2011, 03:41 PM
Pupil

Intersection of a plane and a normal.

Problem: A normal to $3x-y+4z=12$ 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 $3x-y+4z=12$ is $\vec{n} = (3, -1, 4)$ . 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 $x = 2$ and $y = 1$ to get
$3(2) - (1)+4z = 12 => 4z = 7 => z = \frac{7}{4}$ , this point will be $B(2, 1,\frac{7}{4})$ and then find $\vec{AB}$ and take the cross product of $\vec{n}$ and $\vec{AB}$ ?

I just never encountered a question like this before, so I would appreciate some help on getting started. Thanks in advance.
June 4th 2011, 03:49 PM
Plato

Find where the line
$\ell (t) = \left\{ {\begin{array}{*{20}c} {x = 5 + 3t} \\ {y = 1 - t} \\ {z = 6 + 4t} \\ \end{array} } \right.$
intersects the plane.
June 4th 2011, 04:17 PM
Pupil

Quote:

Originally Posted by Plato

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

So, substituting the normal $(3, -1, 4)$ :
$3 = 5 + 3t => -2 = 3t => x = -\frac{2}{3}$
$-1 = 1 - t => y = -2$
$4 = 6 + 4t => -2 = 4t => z = -\frac{1}{2}$

Therefore, the line intersects the plane at the point $(-\frac{2}{3}, -2, -\frac{1}{2}$ , correct?
June 4th 2011, 05:01 PM
HallsofIvy

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.
June 4th 2011, 05:29 PM
Pupil

Ah, I see. I'll reattempt this:
$3(5 + 3t) - (1-t) + 4 (6 + 4t) = 12$
$15 + 9t - 1 + t +24 + 16t = 12$
$26t = -26$
$t = -1$
Substituting this back into the original parametric equation:
$x = 5 + 3(-1) = 2$
$y = 1 - (-1) = 2$
$z = 6 + 4(-1) = 2$
Point of intersection of the normal and the plane is (2, 2, 2). Correct?
June 4th 2011, 05:58 PM
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)?
June 4th 2011, 06:49 PM
Pupil

Quote:

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.