Question

Find an equation of the plane. The plane that passes through the point (−3, 2, 1) and contains the line of intersection of the planes x + y − z = 4 4x − y + 5z = 2

Answer 1

To find the plane's equation, first, find the direction vector by crossing the normals of the given planes, then use this directional vector and the given point (-3, 2, 1) to express the plane's equation in point-normal form.

Step-by-step explanation:

To find an equation of a plane that passes through a point and contains the line of intersection of two given planes, we first need to find the direction vector of the line intersection. To do this, we can take the cross-product of the normal vectors of the given planes. The normal vector for the plane x + y - z = 4 is <1, 1, -1>, and for the plane 4x - y + 5z = 2, it is <4, -1, 5>. The cross product of these two vectors gives us the direction vector for the line of intersection.

Once we have the direction vector (), we use the given point (-3, 2, 1) and the direction vector to express the plane's equation in the form: a(x - x0) + b(y - y0) + c(z - z0) = 0, where () is the point through which the plane passes, and is the direction vector component.