Question

given a plane 2x-3y+5z-10=0 and point (-1,7,4) find a line parallel to the plane and passes through that point.

i know the line equation needs a point and a direction. i have the point. as for the direction, it should be orthogonal to the normal of the plane, which is (2, -3, 5).

two vectors are orthogonal if their dot product is zero. idk how to go from here. help !

Answer 1

This isn't Calculus one is it? xD

Answer 2

(:

Answer 3

http://math.stackexchange.com/questions/477885/find-an-equation-for-the-line-that-is-parallel-to-the-plane-2x-3y-5z-10

there is a fine discussion and solution to your problem here.

good luck

Answer 4

yes i saw that but i am confused at the part where (2,-3,5) . (x,y,z) = 0 like idk how to get xyz. do i just choose any points?

Answer 5

Since 2x−3y+5z−c=0 will be parallel, we can sub in (−1,7,4) to get a c value of c=2(−1)−3(7)+5(4)=−3 , leaving us with the plane:


2x−3y+5z+3=0

Answer 6

follow the along with the bottom answer.

Answer 7

thanks !