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

Nov 8, 2015

Parallel planes have the same perpendicular vector
Multiply the [x,y,z] components of the perpendicular vector into $\left(x - {x}_{0}\right) + \left(y - {y}_{0}\right) + \left(z - {z}_{0}\right) = 0$

Explanation:

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

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:
${\vec{n}}_{\alpha} = \left[- 3 , - 4 , 3\right] = {\vec{n}}_{\beta}$

Now, we have the perpendicular vector of $\beta$ as well.
Next step is to multiply this vector into $\left(x - {P}_{{x}_{0}}\right) \left(y - {P}_{{y}_{0}}\right) \left(z - {P}_{{z}_{0}}\right) = 0$
where ${P}_{{x}_{0}}$ is the x coordinate of the point P and so on.

$\beta = - 3 \left(x - 4\right) - 4 \left(y - \left(- 4\right)\right) + 3 \left(z - \left(- 5\right)\right) = 0$
$\beta = - 3 x + 12 - 4 \left(y + 4\right) + 3 \left(z + 5\right) = 0$
$\beta = - 3 x + 12 - 4 y - 16 + 3 z + 15 = 0$
$\beta = - 3 x - 4 y + 3 z + 11 = 0$

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