Thread: Intersection of a plane and a normal.

Problem: A normal to 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 is . 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 and to get
, this point will be and then find and take the cross product of and ?

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

Find where the line

intersects the plane.

Originally Posted by Plato

Find where the line

intersects the plane.

So, substituting the normal :




Therefore, the line intersects the plane at the point , correct?

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.

Ah, I see. I'll reattempt this:




Substituting this back into the original parametric equation:



Point of intersection of the normal and the plane is (2, 2, 2). Correct?

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)?

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.