The vector perpendicular to the plane is <2,1,-2>.
So the line is x=2t+2, y=t+3, z=-2t-1.
You can insert those values of (x,y,z) into the plane, obtaining t, which will give you the point in (b).
Another one....
Point (P): (2,3,-1)
Plane: 2x + y - 2z + 9 = 0
1. Determine equation of line L, perpindicular to the plane and passing through point P.
2. Determine Point A (where L and plane intersect).
3. Distance from A to P (Point)
4. Also- How would you find the distance from the point using a VECTOR PROJECTION.
5. Knowing that- Find the distance from Point P (-4, 7, -3) to the plane which is 3x + 2y - 5z + 11 = 0.
P.s.- Plato similar type of question but a little different thus giving me some issues.
Thanks a lot!
Joey Colorado.