find the general form of the equation of the plane passing through the point and perpendicular to the specified vector.
(8, 6, 3)
n =i-6j + k
a. x - 6y + z - 25 = 0
b. x - 6y + z + 25 = 0
c. x - 6y + z = 0
d. 8x + 6y + 3z - 25 = 0
e. 8x + 6y + 3z + 25 = 0
Call the vector "u", Then u([x,y,z]-[8,6,3]=0. Simplify to get ans (b).
To find the general form of the equation of the plane passing through the point (8, 6, 3) and perpendicular to the vector n = i - 6j + k, we can use the point-normal form of the equation of a plane.
First, let's find the components of the normal vector to the plane:
n = i - 6j + k
The components of the normal vector are A = 1, B = -6, and C = 1.
Next, substitute the coordinates of the given point (8, 6, 3) into the equation of the plane:
A(x - x0) + B(y - y0) + C(z - z0) = 0
(1)(x - 8) + (-6)(y - 6) + (1)(z - 3) = 0
x - 8 - 6y + 36 + z - 3 = 0
x - 6y + z + 25 = 0
So, the correct answer is option b. x - 6y + z + 25 = 0.
To find the general form of the equation of a plane passing through a specific point and perpendicular to a given vector, you can use the point-normal form of the equation of a plane.
The point-normal form of the equation of a plane is given by:
Ax + By + Cz + D = 0
Where (A, B, C) are the direction ratios of the vector perpendicular to the plane, and (x, y, z) are the coordinates of any point on the plane.
In this case, the given point is (8, 6, 3), and the given vector is n = i - 6j + k.
To determine the direction ratios of the normal vector, we compare it with the general form of a vector, r = xi + yj + zk. So, the direction ratios are A = 1, B = -6, C = 1.
Substituting the values of A, B, and C along with the coordinates of the point (8, 6, 3) into the point-normal form, we get:
1(8) - 6(6) + 1(3) + D = 0
8 - 36 + 3 + D = 0
-25 + D = 0
D = 25
Therefore, the general form of the equation of the plane passing through the point (8, 6, 3) and perpendicular to the given vector n = i - 6j + k is:
x - 6y + z - 25 = 0.
So, the correct answer is option a) x - 6y + z - 25 = 0.
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