Question

Find the equation for the plane containing the point P(2,5, -6) and parallel to the plane : 2x + 5y +7z=12.

A. X-2/1 = y-5/4 = 2-7/-6
B. (x - 2)+4(y-5)-6(z-7) = 0
C. x - 2/2 = y - 5/5 = z + 6/7
D. 2(x-2)+5(y-5)+7(2+6)=0

Answer 1

The equation of the plane containing the point P(2,5,-6) and parallel to the plane 2x + 5y +7z = 12 is 2(x - 2) + 5(y - 5) + 7(z + 6) = 0.

Step-by-step explanation:

The student is asking for the equation of a plane that passes through the point P(2,5,-6) and is parallel to the given plane 2x + 5y +7z = 12. Since the planes are parallel, they will have the same normal vector, which is given by the coefficients of x, y, and z in the equation of the given plane. Therefore, the normal vector is (2, 5, 7).

We can use the point-normal form of the equation of a plane, which is given by: A(x - x0) + B(y - y0) + C(z - z0) = 0, where (A,B,C) is the normal vector and (x0, y0, z0) is a point on the plane. Using the normal vector (2, 5, 7) and the point P(2, 5, -6), we get:

2(x - 2) + 5(y - 5) + 7(z + 6) = 0.