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 $C}_{2$

And find the other plane

Karen Robbins

Beginner2021-12-28Added 49 answers

Well, we can rewrite ${C}_{2}:3x+3y+6z=18$

Two plain have normal${n}^{\to}=[3,3,6]$

Chose$P,Q$ from $C}_{1},{C}_{2$ respectively.

Say,$P(5,0,0)$ and $Q(-\frac{1}{2},0,0)$

Then,$Q{P}^{\to}=[-\frac{11}{2},0,0]$

$Proj\left\{{}_{n}^{\to}\right\}Q{P}^{\to}=\frac{Q{P}^{\to}{n}^{\to}}{{\left|\left|{n}^{\to}\right|\right|}^{2}}\times {n}^{\to}=\frac{-\frac{33}{2}}{q+q+36}[3,3,6]$

$=-\frac{11}{36}[3,3,6]=[-\frac{11}{12},-\frac{11}{12},-\frac{11}{6}]$

Thus, the distance is$\left|\left|Proj\left\{{}_{n}^{\to}\right\}Q{P}^{\to}\right|\right|=\sqrt{{\left(\frac{11}{12}\right)}^{2}+{\left(\frac{11}{12}\right)}^{2}+{\left(\frac{11}{6}\right)}^{2}}=\sqrt{\frac{121}{24}}$

Two plain have normal

Chose

Say,

Then,

Thus, the distance is

Bernard Lacey

Beginner2021-12-29Added 30 answers

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 $\frac{(1,1,2)}{\sqrt{1+1+4}}=\frac{(3,3,6)}{\sqrt{9+9+36}}$ , which means these two planes are parallel and we can write the normal as $\frac{1}{\sqrt{6}}(1,1,2)$ .

Now we should find two points on the planes.

Let$y=0$ and $z=0$ , and find the corresponding $x$ values. For ${C}_{1},\text{}x=4$ and for ${C}_{2},\text{}x=6$ . So we know $C}_{1$ contains the point $(4,0,0)$ and $C}_{2$ contains the point $(6,0,0)$ .

The distance between these points is$2$ and the direction is $(1,0,0)$ . We now that this is not the shortest distance between these two points as $(1,0,0)\ne \frac{1}{\sqrt{6}}(1,1,2)$ so the direction is not perpendicular to these planes. But this is ok as we can use the dot product between $(1,0,0)$ and $\frac{1}{\sqrt{6}}(1,1,2)$ to work out the proportion of the distance that is perpendicular to the planes.

$(1,0,0)\times \frac{1}{\sqrt{6}}(1,1,2)=\frac{1}{\sqrt{6}}$

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, $\frac{1}{\sqrt{6}}(1,1,2)$ . Using this we can write ${C}_{3}:x+y+2z=a$ and determine $a$ . When $y=0,z=0$ we moved from $(4,0,0)\to (6,0,0)$ , so if we move the same distance again we go $(6,0,0)\to (8,0,0)$ and $(8,0,0)$ is on $C}_{3$ .

Therefore,$a=8$ . So final equation of the plane is: ${C}_{3}:x+y+2z=8$

Now we should find two points on the planes.

Let

The distance between these points is

So the distance between the two planes is

The last part is to find the plane which is the same distance away from

Therefore,

nick1337

Expert2022-01-08Added 777 answers

Thanks a lot!

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