Find a scalar equation of the plane that contains the given point P=(-3,-3,1)and the given normal vector n = (-1,4,7)
To find the scalar equation of the plane, we first need to understand the general form of the equation of a plane.
The equation of a plane can be written as Ax + By + Cz + D = 0, where A, B, and C are the coefficients corresponding to the variables x, y, and z respectively, and D is the constant term.
In this case, we are given a point P=(-3,-3,1) on the plane and a normal vector n = (-1,4,7) that is perpendicular to the plane.
To find the coefficients A, B, C, and D in the equation of the plane, we can use the following steps:
Step 1: Use the coordinates of the point P=(-3,-3,1) to substitute into the equation of the plane:
A(-3) + B(-3) + C(1) + D = 0
Step 2: Use the components of the normal vector n=(-1,4,7) to determine the coefficients A, B, and C. Since the normal vector is perpendicular to the plane, its components correspond to the coefficients in the equation of the plane:
A = -1, B = 4, C = 7
Step 3: Substitute the values of A, B, C into the equation from Step 1:
-1(-3) + 4(-3) + 7(1) + D = 0
Simplifying this equation, we get:
3 - 12 + 7 + D = 0
-2 + D = 0
Step 4: Solve for D:
D = 2
Therefore, the scalar equation of the plane that contains the point P=(-3,-3,1) and the normal vector n=(-1,4,7) is:
-1x + 4y + 7z + 2 = 0