# Thread: Finding an equation for the plane....

1. ## Finding an equation for the plane....

I am unsure of how to solve this linear algebra problem...

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

Help would be greatly appreciated!

2. The normal vectors to the planes $\displaystyle x - y - 4z = 2$ and $\displaystyle -2x + y + 2z = 3$ are $\displaystyle <1, -1, -4>$ and $\displaystyle <-2, 1, 2>$. If you take the cross-product of these two vectors, you will get a vector that points in the same direction of the line. Then, all you have to do is find two points A and B on that line, and let C = $\displaystyle (2, 4, -1)$. Then the cross-product of the vectors AC and BC will be a normal vector $\displaystyle <p, q, r>$ to the desired plane, whose equation will be $\displaystyle px + qy + rz = s$, and s can be determined by putting point C into the equation: $\displaystyle 2p + 4q - r = s$.

3. how can you find the equation of the plane in ax+by+cz=d format through the point (1,2,3) for example with normal vector (2,-1,4)
thanks for any help

4. sorry just dont understand wher yo use the normal vector after gettin x+y+z= s