Find the general form of the equation of the plane passing through the point and perpendicular to the specified vector.[Be sure to reduce the coefficients in your answer to lowest terms by dividing out any common factor.] (6,2,9),n = 8i + 6j + 9k
A)8x + 6y + 9 z = 0
B)8x + 6y + 9 z + 141 = 0
C)8x + 6y + 9 z - 141 = 0
D)6x + 2y + 9z + 141 = 0
E)6x + 2y + 9z - 141 = 0

Question

Quizplus · Accepted Answer

The equation of a plane can be written as ax + by + cz + d = 0, where (a,b,c) is the normal vector to the plane and (x,y,z) is any point on the plane. 
In this case, the given point is (6,2,9) and the normal vector is 8i + 6j + 9k. 
Substituting these values into the equation, we get:
8x + 6y + 9z - d = 0
To find d, we plug in the coordinates of the given point:
8(6) + 6(2) + 9(9) - d = 0
Simplifying, we get:
48 + 12 + 81 - d = 0
d = 141
Substituting this value back into the equation, we get:
8x + 6y + 9z - 141 = 0
Reducing the coefficients to lowest terms, we get:
Answer: C
 The equation of a plane can be written as ax + by + cz + d = 0, where (a,b,c) is the normal vector to the plane and (x,y,z) is any point on the plane. In this case, the given point is (6,2,9) and the normal vector is 8i + 6j + 9k. Substituting these values into the equation, we get 8x + 6y + 9z - d = 0. To find d, we plug in the coordinates of the given point, and we get d = 141. Substituting this value back into the equation, we get 8x + 6y + 9z - 141 = 0. Finally, reducing the coefficients to lowest terms, we get the final answer of choice C.

Find the General Form of the Equation of the Plane