Thread: Equation of a plane

This is a question from my calculus 3 class.

"Find an equation of the plane through the point (-1,4,2) that contains the line of intersection of the planes (4x - y + z - 2) = 0 and (2x + y - 2z - 3) = 0."

I'm having problems trying to figure this out. I get an answer but it's different than what the book says it is.

My work:
v1 = <4,-1,1>
v2 = <2,1,-2>

v1 x v2 = i + 10j + 6k

Point-normal form equation of plane:
(x+1) + 10(y-4) + 6(z-2) = 0
x + 1 + 10y - 40 + 6z - 12 = 0
x + 10y + 6z = 51 <---- my answer?

The book says the answer is 4x - 13y + 21z = -14, and more than likely it is correct and not me in this case.

Step 1: Find the line of intersection of the planes.

Step 2: Find two points on this line.

Step 3: The two points on your line that you found in step 2, along with the point (-1, 4, 2), can be used to find the equation of the plane.

Cool, I'm going to do that.