Question

Find an equation of the plane that passes through the point (-2, 3, 1) and contains the line of intersection of the planes xy - z = 5 and 2x - y + 3z = 2?

Answer 1

To find the equation of the plane, we first find the normal vector to the plane by taking the cross product of the direction vectors of the two given planes. Then, using the point-normal form of the equation of a plane, we can find the equation by plugging in the point and the normal vector. The equation of the plane is -4x -y -3z -5 = 0.

Step-by-step explanation:

To find the equation of the plane, we need to find the normal vector to the plane. The normal vector can be found by taking the cross product of the direction vectors of the two given planes. We can then use the point-normal form of the equation of a plane to find the equation. First, let's find the direction vectors of the two planes:

1. For the plane xy - z = 5, the direction vector is <1, 1, -1>
2. For the plane 2x - y + 3z = 2, the direction vector is <2, -1, 3>

Taking the cross product of the two direction vectors, we get:

<1, 1, -1> x <2, -1, 3> = <-4, -1, -3>

Now, let's find the equation of the plane using the given point and the normal vector:

The equation of the plane is -4(x - (-2)) -1(y - 3) -3(z - 1) = 0 simplified to -4x -y -3z -5 = 0.