Question

How to find an equation of the plane through the point (4, -4, -5) and parallel to the plane −3x−4y+3z=−5 ?

Answer 1

Parallel planes have the same perpendicular vector
Multiply the [x,y,z] components of the perpendicular vector into `(x-x_0) + (y-y_0) + (z-z_0) = 0`

Explanation:

First of all, let's name the planes `alpha` and `beta` .
`alpha = -3x - 4y + 3z = -5`
`beta` is parallel with `alpha` ang goes through `P(4, -4, -5)`

When we have the equation of a plane, we have the perpendicular vector as well! You can tell what the perpendicular vector is by looking at the coefficients of the x, y and z components of the plane..

In this case, the perpendicular vector of `alpha` will be:
`vec(n)_alpha = [-3, -4, 3]= vec(n)_beta`

Now, we have the perpendicular vector of `beta` as well.
Next step is to multiply this vector into `(x-P_(x_0))(y-P_(y_0))(z-P_(z_0))=0`
where `P_(x_0)` is the x coordinate of the point P and so on.

`beta = -3(x-4) - 4(y - (-4)) + 3(z-(-5)) = 0`
`beta = -3x + 12 -4(y+4) + 3(z+5) = 0`
`beta = -3x + 12 -4y - 16 +3z +15 = 0`
`beta = -3x - 4y + 3z + 11 = 0`

Notice: The only thing that changed is`d` (which is the constant in the equation). Remember that parallel planes have the same perpendicular vector!