Find an equation of the plane through the point and perpendicular to the given line?

Find an equation of the plane that passes through the point an contains the line of intersection?

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

• Answer:

  The line of intersection will be included in both planes so that it will be orthogonal to both their normals so find the cross product of the normals. <1,1,-1>×<3,-1,4> = <1(4) - (-1)(-1), -((1)(4) - 3(-1)), (1)(-1) - (1)(3)> = <3,-7,-4> So that's the direction vector for the line and to find a point in the line eliminate one of the variables by adding the two equations together. 4x + 3z = 4 Now just pick z = 0 x = 1 So (1,0,0) is in the line. To find the plane containing the line and the point make a vector out of this point and the given point and find the cross product (yea I know I don't like cross products either) between this vector and the direction vector of the line. That will be the normal for the plane. <-1 - 1, 1 - 0, 3 - 0> = <-2,1,3> <-2,1,3>×<3,-7,-4> = <17,1,11> Now use the normal and the point to find the equation of the line, you can use (1,0,0), it's easier. 17(x - 1) + y + 11(z - 0) = 0 17x + y + 11z = 17 Not a bad problem, little confusing, but all in all not too horrible.