How do you find the parametric equation through the point (6, 0, 10) and parallel to the plane -3x+2y-4z = 0?

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Answer 1 · 2016-12-14T20:39:23

Please see the explanation.

Any plane parallel to the plane #-3x + 2y - 4z = 0# will have an equation of the same form but equal to a different constant:

#-3x + 2y - 4z = C" [1]"#

To force the point #(6, 0,10)# to be in the plane, substitute the point into equation [1] and solve for C:

#-3(6) + 2(0) - 4(10) = C#

#C = -58#

The scalar equation of the plane parallel to the given plane is:

#-3x + 2y - 4z = -58" [2]"#

The vector equation of the plane is

#(x, y ,z) = (6, 0, 10) + s(baru) + t(barv)#

Where #baru and barv# are any two vectors in the plane.

Let's use equation [2] to find two other points in the plane:

Let #x = 6 and y = 2#

#-3(6) + 2(2) - 4z = -58#

#-4z = -44#

#z = 11#

The vector from the point #(6, 0, 10)# to the point #(6,2, 11)# is:

#baru = 2hatj + hatk#

Let #x = 10 and y = 0#

#-3(10) + 2(0) - 4z = -58#

#-4z = -28#

#z = 7#

The vector from point #(6, 0, 10)# to the point #(6, 0, 7)# is:

#barv = 4hati - 3hatk#

A vector equation of the plane is:

#(x, y ,z) = (6, 0, 10) + s(2hatj + hatk) + t(4hati - 3hatk)#

From the above equation, we can write a set of parametric equations:

#x = 4t + 6#
#y = 2s#
#z = s - 3t + 10#

NOTE: These are not unique equations, because we could have chosen other points and obtained different vectors.