Consider the line and plane below.
x = 5-5t, y = 3+7t, z = 10t
ax + by + cz = d
Find values of a, b, c, and d so that the plane is perpendicular to the line and through the point (2, 1, 2).
Fgrad=(x',y',z') is perp to surface
The Attempt at a Solution
when I put those x,y,z values together I get a parametric equation that equals <5,3,0>+t<-5,7,10>
the starting point <5,3,0> I think is not relevant to finding a perpendicular vector,
I just need to find a vector perpendicular to t<-5,7,10> that goes through (2,1,2) i think? but i'm not sure how to do this..
then once I find the vector thats perp to t<-5,7,10>(dot)<a,b,c>=0
once I know a,b,c I can solve for d