Question

Need Calculus assistance. Find the equation of the plane containing the point (5,-1,0) and perpendicular to the line x=2t,y=t+3, z=3t-1?

Answer 1

I assume that my n vector for the line is <2,1,3> and that since 0 equals the dot product of n1 x n2 if I let n1 equal <2,1,3> then n2 can be any plane as long as long as when multiplied by <2,1,3> equals 0. Is this correct assumption?

Answer 2

0 would be the dot product of two orthogonal lines is what I mean.

Answer 3

yes normal vector would be perpendicualr to every vector in the plane

Answer 4

suppose \((x,y,z)\) is any point on the plane, then we have
\[(x-5,y+1,z-0) \bullet (2,1,3) = 0\]

Answer 5

ii see (x,y,z) represent a second point on the plane and (x-5,y+1, z-0) would represent the vector and the dot product with the orthogonal vector <2,1,3> we arrive at zero.

Answer 6

thats right!

Answer 7

Thank you

Answer 8

yw