Question

Find the equation of the plane through (1,2,-2) that contains the line x=2t, y=3-t z=1+3t

Answer 1

The line passes through the point B=(0,3,1) and parallelto the vector a=(2,-1,3), so the equation of the point on that line can be
written as follows
X(t) = B + a t The plane is parallel to vectors
a=(2,-1,3)
and
AB=(0-1, 3-2,1-(-2)) = (-1,1,3) Hence the normal vector n=(nx,ny,nz) to p is a crossproduct of a and AB.
Thus nx = | -1 3 | = -1*3 -1*3 = -6
| 1 3 |ny = | 3 2 | =3*(-1) - 3*2 = -9
| 3 -1 | nz = | 2 -1 | = 2- (-1)*(-1) = 2-1=1
| -1 1 | So n = (-6,-9,1)
Therefore the plane p is given by the following equationof the plane passing through point A=(1,2,-2) and orthogonal to the vector
n=(-6,-9,1): -6(x-1) -9(y-2) + z-2=0
-6x + 6 -9y + 18 + z-2=0
- 6x - 9y + z + 22 = 0