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Write an equation of the plane with normal vector n =⟨−7,−4,−6⟩ passing through the point (−8,4,6) in scalar form in the form of equation = 20

Question

Write an equation of the plane with normal vector n=7,4,6\mathbf{n} = \langle -7, -4, -6 \rangle

passing through the point (8,4,6)(-8, 4, 6) in scalar form
in the form of equation =20= 20

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Solution

The equation of a plane in 3D space is given by the formula:

Ax + By + Cz = D

where A, B, and C are the components of the normal vector to the plane, and D is a constant. The point (x0, y0, z0) lies on the plane.

Given the normal vector n = ⟨−7,−4,−6⟩ and the point (−8,4,6), we can substitute these values into the equation:

-7(x - (-8)) - 4(y - 4) - 6(z - 6) = 0

Simplify this to get:

-7x + 56 - 4y + 16 - 6z + 36 = 0

Combine like terms to get:

-7x - 4y - 6z + 108 = 0

We want the equation in the form of equation = 20, so subtract 108 from both sides to get:

-7x - 4y - 6z = -88

This is the equation of the plane in scalar form.

This problem has been solved

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