We know one direction vector of the plane is (-2,3,1)
We have to find another.
Use the two points (3,5,1) and (1,2,3)
the vector would be (2,3,-2)
I will assume you know how to find the cross-product between these two vectors.
I got (9,2,12)
So the plane has equation
9x + 2y + 12z = c
plug in (1,2,3)
9 + 4 + 36 = c = 49
scalar equation is
9x + 2y + 12z = 49
vector equation is
r = (3,5,1) + t(-2,3,1) + s(2,3,-2)
Find the vector and scalar equations of the plane containing the line r=(3,5,1)+t(-2,3,1) and the point (1,2,3)
1 answer
1 answer