Question

Find an equation for the plane containing point P, parallel to vector A, and orthogonal to the plane found in (c). Write all three forms of the equation for the plane.

a) (Ax + By + Cz = D)
b) (A(x - x_0) + B(y - y_0) + C(z - z_0) = 0)
c) (A(x - x_0) + B(y - y_0) + C(z - z_0) = D)
d) (Ax + By + Cz + D = 0)

Answer 1

To find the plane's equation, identify the normal vector and utilize the point P's coordinates. Apply the equation in the general form (Ax + By + Cz = D), adjusting for point P. Ensure the normal vector is not parallel to vector A.

Step-by-step explanation:

To find the equation of a plane containing point P and parallel to vector A, we first need to identify the normal vector to the plane. Since the plane must be orthogonal to a given plane, the normal vector of the second plane can serve as the direction vector for our target plane. Let's assume the normal vector is of the form (A, B, C), and the point P has coordinates (x0, y0, z0).

The general equation for the plane in the different forms requested is as follows:

• (a) Ax + By + Cz = D
• (b) A(x - x0) + B(y - y0) + C(z - z0) = 0
• (c) A(x - x0) + B(y - y0) + C(z - z0) = D
• (d) Ax + By + Cz + D = 0

Where D is the dot product of the normal vector with point P, resulting in D = Ax0 + By0 + Cz0. In converting from vector components to a plane equation, analytical methods of vector addition and subtraction are important. Given that the vector A is parallel to the plane, we must ensure that the normal vector is not parallel to A.