What is the distance between the planes #2x – 3y + 3z = 12# and #–6x + 9y – 9z = 27#?

Question

What is the distance between the planes #2x – 3y + 3z = 12#  and #–6x + 9y – 9z = 27#?

Wyatt Anderson · Accepted Answer

Distance between the planes #2x-3y+3z=12# and #-6x+9y-9z=27# is #4.48#.

First let us find a point on one plane. For this in plane #2x-3y+3z=12#, assume #x=y=0#, then we have #3z=12# or #z=4# and hence #(0,0,4)# is on this plane. Now we find the distance between point #(0,0,4)# and plane #-6x+9y-9z=27# or #-6x+9y-9z-27=0#. As distance from a point #(x_1,y_1,z_1)# to plane #ax+by+cz+d=0# is #D=|ax_1+by_1+cz_1+d|/sqrt(a^2+b^2+c^2)# The distance from point #(0,0,4)# to plane #-6x+9y-9z-27=0# is given by #D=|-6xx0+9xx0-9xx4-27|/sqrt(6^2+9^2+9^2)=|-36-27|/sqrt(36+81+81)# or #D=63/sqrt(198)=63/(3sqrt22)=21/sqrt22=4.48#

Claire Akin · Answer

undefined To find the distance between two parallel planes given by their equations, you can use the formula:

$$ \text{Distance} = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} $$

Where $A$, $B$, and $C$ are the coefficients of the variables $x$, $y$, and $z$ in the equations of the planes, and $D_1$ and $D_2$ are the constants on the right side of the equations of the planes.

For the planes $2x - 3y + 3z = 12$ and $-6x + 9y - 9z = 27$, the coefficients are $A_1 = 2$, $B_1 = -3$, $C_1 = 3$ for the first plane, and $A_2 = -6$, $B_2 = 9$, $C_2 = -9$ for the second plane. The constants are $D_1 = 12$ and $D_2 = 27$.

Plugging these values into the formula, we get:

$$ \text{Distance} = \frac{|12 - 27|}{\sqrt{2^2 + (-3)^2 + 3^2}} $$

$$ = \frac{|-15|}{\sqrt{4 + 9 + 9}} $$

$$ = \frac{15}{\sqrt{22}} $$

$$ = \frac{15\sqrt{22}}{22} $$

So, the distance between the planes is $ \frac{15\sqrt{22}}{22} $.