Find where the line
intersects the plane.
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.
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.
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)?