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

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Nov 8, 2015

Answer:

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!

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