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.]
(8, 6, 3), n = i – 6j + k
To find the equation of a plane passing through a given point and perpendicular to a given vector, there are a few steps we need to follow:
1. Obtain the coordinates of the given point: In this case, the point is (8, 6, 3).
2. Write the equation of the plane in the form ax + by + cz + d = 0, where (a, b, c) is the normal vector of the plane and d is a constant.
3. Determine the normal vector of the plane: In this case, the given vector is n = i - 6j + k. Since the normal vector of the plane is perpendicular to the plane itself, it is also perpendicular to the given vector.
4. Substitute the coordinates of the given point and the normal vector into the equation of the plane to determine the constant d.
Let's go through the steps in detail:
Step 1: Given point (8, 6, 3).
Step 2: Equation of the plane: ax + by + cz + d = 0.
Step 3: Normal vector of the plane: n = i - 6j + k.
Step 4: Substitute the coordinates of the given point and normal vector.
Plugging in the values:
a * 8 + b * 6 + c * 3 + d = 0.
Now, to determine the coefficients a, b, c, and the constant d, we need to simplify this equation by equating the coefficients of the unit vectors i, j, and k on both sides of the equation.
Equating the coefficients of i:
8a + 6b + 3c = 0.
Equating the coefficients of j:
0i - 6b + 0c = 0.
-6b = 0.
b = 0.
Equating the coefficients of k:
0i + 0b + 1c = 0.
c = 0.
Now, we can substitute b = 0 and c = 0 into the equation to solve for a and d.
8a + 6(0) + 3(0) = 0.
8a = 0.
a = 0.
Therefore, a = 0, b = 0, and c = 0.
The equation of the plane passing through the point (8, 6, 3) and perpendicular to the vector n = i - 6j + k is:
0x + 0y + 0z + d = 0.
d = 0.
So, the equation of the plane is simply 0 = 0.