Find an equation for the plane that contains the line v = (-1,1,2) + t(3,2,4) and is perpendicular to the plane 2x + y -3z + 4 = 0
So I solve the line v to get the set of parametric equations
x = -1 + 3t
y = 1 +2t
z = 2 +4t
Then I subsituted them into the equation for the plane 2x + y -3z + 4 = 0
2(-1 + 3t) + (1 + 2t) - 3(2 + 4t) = -4
Solving for t
t = -3/4
I put this value for t back into my set of parametric equations to solve for x,y,z to get a point P
P = (-13/4, -1/2, -1)
I then put this into point normal form and compute the dot product to get the standard equation for the plane
n = (2, 1, -3) P = P = (-13/4, -1/2, -1)
(2, 1, -3) . (x - (-13/4), y - (-1/2), z - (-1)) = 0
2x + 13/2 + y + 1/2 - 3z - 3 = 0
2x + y -3z = -4
So now I'm lost because I just go right back to where I started with the original equation for the plane. What am I doing wrong?
So I solve the line v to get the set of parametric equations
x = -1 + 3t
y = 1 +2t
z = 2 +4t
Then I subsituted them into the equation for the plane 2x + y -3z + 4 = 0
2(-1 + 3t) + (1 + 2t) - 3(2 + 4t) = -4
Solving for t
t = -3/4
I put this value for t back into my set of parametric equations to solve for x,y,z to get a point P
P = (-13/4, -1/2, -1)
I then put this into point normal form and compute the dot product to get the standard equation for the plane
n = (2, 1, -3) P = P = (-13/4, -1/2, -1)
(2, 1, -3) . (x - (-13/4), y - (-1/2), z - (-1)) = 0
2x + 13/2 + y + 1/2 - 3z - 3 = 0
2x + y -3z = -4
So now I'm lost because I just go right back to where I started with the original equation for the plane. What am I doing wrong?