I'm studying for my linear algebra test tomorrow and came across the following problem:
Find all the points on e which are located on a distance sqrt26 from the plane a.
e: x/2 = (y-5)/3 = z-2
A is the plane through:
The point Q(0,10,1)
And the line (let's call it f) :
3x-4z+9 = 0
7z-3y+3 = 0
This was my idea:
- Find the direction vector for f by the cross product off the normal vectors of the planes which define f.
- Find the normal vector of a perpendicular to f.
- Define the plane with the point Q and the normal vector
- Determine the distance from the plane to the line a with the distance formula.
I already went wrong on the second step, i can't find the normal vector for the plane. Does anybody have any ideas for that?
Also, if there is another way of solving this problem, please tell me.
Any help at all is very much appreciated!