Final answer:
To find the equation of the plane containing both the point (-2, 1, 2) and the line of intersection of given planes, the cross product of the planes' normals provides the direction vector for the plane, which, along with the point, establishes the desired plane's equation.
Step-by-step explanation:
The student is asking for the equation of a plane that not only passes through a given point (-2, 1, 2) but also contains the line of intersection of two given planes with equations xy - z = 5 and 3x - y + 5z = 4. To find the equation of such a plane, we essentially look for a plane that encompasses both the point and the line where the two given planes intersect. Assuming that the point lies on the desired plane, its normal vector should be perpendicular to the normal vectors of the given planes.
The normal vector of the plane xy - z = 5 is (1, -1, -1) and the normal vector of the plane 3x - y + 5z = 4 is (3, -1, 5). The normal vector of the desired plane is also the cross product of these two normals. After we find this vector, we can use it along with the point (-2, 1, 2) to write the equation of the plane.
Cross product calculation
:
- Let ν1 = (1, -1, -1)
- Let ν2 = (3, -1, 5)
- The cross product ν1 × ν2 will give the normal vector N of the desired plane.
Once we have the normal vector N, we use the point (-2, 1, 2) and the normal vector to write the plane equation in the form: N · (x - x0, y - y0, z - z0) = 0, where (x0, y0, z0) is the point (-2, 1, 2).