Distance Between 2 Planes: 6z = 4y - 2x, 9z = 1 - 3x + 6y

[jks-14]

Hi,
I have a problem which i am struggling to solve...

Qu: Find the distance between the planes 6*z=4*y-2*x and 9*z=1-3*x+6*y.

I've rearranged the equation and everything to obtain
P1 : -2*x+4*y-6*z=0
P2 : 3*x-6*y+9*z=1

V1 = <-2,4,-6>
V2 = <3,-6,9>

But this then means that the normal vector equals <0,0,0> and i also wasn't sure which point i would then use to calculate the distance between.

I was wondering if maybe the way to solve the question was to choose 2 points in each plane and solve from there but wasn't 100% sure how i would approach the question in such a way.

Thanks for any help... Much appreciated!

 

[soroban]

Re: Distance Between 2 Plance

Hello, jks-14!

There is a formula for this problem, but I'll walk you through the solution.

\(\displaystyle \text{Find the distance between the planes: }\;\begin{array}{c}6z\:=\:4y-2x \\ 9z\:=\:1-3x+6y\end{array}\)

\(\displaystyle \text{We have: }\;\begin{array}{cccc}2x-4y + 6z &=& 0 & [1] \\ 3x - 6y + 9z &=& 1 & [2]\end{array}\)

We want the perpendicular distance between the two planes.

\(\displaystyle \text{The vectors perpendicular to the planes are: }\:\langle2,\text{-}4,6\rangle\:\text{ and }\:\langle 3,\text{-}6,9\rangle\)
. . \(\displaystyle \text{which are equal to: }\:\langle 1,\text{-}2,3\rangle\)


\(\displaystyle \text{Pick any point on plane [1], say: }\(0,0,0)\)

\(\displaystyle \text{The line through }(0,0,0)\text{ with direction }\langle1,\text{-}2,3\rangle\:\text{ has equations: }\:\begin{Bmatrix}x &=& t \\ y &=& \text{-}2t \\ z &=& 3t \end{Bmatrix}\;\;[3]\)


\(\displaystyle \text{Where does this line intersect plane [2]?}\)

\(\displaystyle \text{Substitute [3] into [2]: }\quad 3(t) - 6(-2t) + 9(3t) \:=\:1 \quad\Rightarrow\quad 42t \:=\:1 \quad\Rightarrow\quad t \:=\:\frac{1}{42}\)

\(\displaystyle \text{Substitute into [3]: }\;x =\frac{1}{42},\;\;y =-\frac{2}{42},\;\;z =\frac{3}{42}\)

. . \(\displaystyle \text{Hence: }\;Q\left(\frac{1}{42},\:\text{-}\frac{2}{42},\:\frac{3}{42}\right)\)


\(\displaystyle \text{Now use the Distance Formula to find the distance }\overline{PQ}.\)