The cross-product of the two given vectors will give us the normal that we can use for the new plane.
You should have a method of your own to find that normal, I had (1,0,0)
(notice the dotproduct of (1,0,0) with each of the given vectors is zero)
so the equation of the new plane is
x + 0y + 0z = c
but the point (0,6,-2) lies on it, so
0 + 0 + 0 = c
c = 0
the equation of the plane is x = 0
find an equation of a plane containing a point (0,6,-2) and parallel to vector v1= (0,-3,0), v2=(0,9,-1)
Help!
2 answers
find a line that passes through point (2,5,-3) and is perpendicular to the plane 2x-3y+4z+7=0
2 answers