Michael Maggard

2021-12-27

Find the distance between two planes:
${C}_{1}:x+y+2z=4$ and ${C}_{2}:3x+3y+6z=18$
And find the other plane ${C}_{3}\ne {C}_{1}$ that has the distance d po the plain ${C}_{2}$

Karen Robbins

Expert

Well, we can rewrite ${C}_{2}:3x+3y+6z=18$
Two plain have normal ${n}^{\to }=\left[3,3,6\right]$
Chose $P,Q$ from ${C}_{1},{C}_{2}$ respectively.
Say, $P\left(5,0,0\right)$ and $Q\left(-\frac{1}{2},0,0\right)$
Then, $Q{P}^{\to }=\left[-\frac{11}{2},0,0\right]$
$Proj\left\{{}_{n}^{\to }\right\}Q{P}^{\to }=\frac{Q{P}^{\to }{n}^{\to }}{{||{n}^{\to }||}^{2}}×{n}^{\to }=\frac{-\frac{33}{2}}{q+q+36}\left[3,3,6\right]$
$=-\frac{11}{36}\left[3,3,6\right]=\left[-\frac{11}{12},-\frac{11}{12},-\frac{11}{6}\right]$
Thus, the distance is $||Proj\left\{{}_{n}^{\to }\right\}Q{P}^{\to }||=\sqrt{{\left(\frac{11}{12}\right)}^{2}+{\left(\frac{11}{12}\right)}^{2}+{\left(\frac{11}{6}\right)}^{2}}=\sqrt{\frac{121}{24}}$

Bernard Lacey

Expert

For a plane defined by $ax+by+cz=d$ the normal is said to be (a,b,c). This is a direction, so we can normalise it $\frac{\left(1,1,2\right)}{\sqrt{1+1+4}}=\frac{\left(3,3,6\right)}{\sqrt{9+9+36}}$, which means these two planes are parallel and we can write the normal as $\frac{1}{\sqrt{6}}\left(1,1,2\right)$.
Now we should find two points on the planes.
Let $y=0$ and $z=0$, and find the corresponding $x$ values. For and for . So we know ${C}_{1}$ contains the point $\left(4,0,0\right)$ and ${C}_{2}$ contains the point $\left(6,0,0\right)$.
The distance between these points is $2$ and the direction is $\left(1,0,0\right)$. We now that this is not the shortest distance between these two points as $\left(1,0,0\right)\ne \frac{1}{\sqrt{6}}\left(1,1,2\right)$ so the direction is not perpendicular to these planes. But this is ok as we can use the dot product between $\left(1,0,0\right)$ and $\frac{1}{\sqrt{6}}\left(1,1,2\right)$ to work out the proportion of the distance that is perpendicular to the planes.
$\left(1,0,0\right)×\frac{1}{\sqrt{6}}\left(1,1,2\right)=\frac{1}{\sqrt{6}}$
So the distance between the two planes is $\frac{2}{\sqrt{6}}$.
The last part is to find the plane which is the same distance away from ${C}_{2}$ as ${C}_{1}$ but in the opposite direction. We know the normal must be the same, $\frac{1}{\sqrt{6}}\left(1,1,2\right)$. Using this we can write ${C}_{3}:x+y+2z=a$ and determine $a$. When $y=0,z=0$ we moved from $\left(4,0,0\right)\to \left(6,0,0\right)$, so if we move the same distance again we go $\left(6,0,0\right)\to \left(8,0,0\right)$ and $\left(8,0,0\right)$ is on ${C}_{3}$.
Therefore, $a=8$. So final equation of the plane is: ${C}_{3}:x+y+2z=8$

nick1337

Expert