Posted by **Amber** on Monday, June 10, 2013 at 3:32pm.

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 x-y-2z+3=0 and 3x-2y+z+5=0

- Calculus Grade 12 -
**Steve**, Monday, June 10, 2013 at 4:03pm
The line of intersection of the planes is in the direction of the cross-product 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(x-3) - 7(y-7) + 1(z-1) = 0

-5x + 15 - 7y + 49 + z - 1 = 0

5x+7y-z = 63

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