Calculus Grade 12
posted by Amber on .
Find the equation of the plane that passes through the point (3,7,1) and is perpendicular to the line of intersection of the planes xy2z+3=0 and 3x2y+z+5=0

The line of intersection of the planes is in the direction of the crossproduct of the two normals:
{1,1,2}x{3,2,1} = {5,7,1}
Now we have a point and a normal vector.
The equation of the plane is thus
5(x3)  7(y7) + 1(z1) = 0
5x + 15  7y + 49 + z  1 = 0
5x+7yz = 63