$3$d, geometry, planes.

Question

In a question, we are asked to

Find a plane in $\Bbb R^3$ that is perpendicular to each of the planes: $2x+3y-2z=5$ and $x+2y-3z=8$.

My question is: if this third plane is supposed to be perpendicular to both of those planes, shouldn't both of those two planes be either mutually perpendicular or parallel? If we find the dot product of the normal vectors of those planes, it isn't $0$, and the normal vectors aren't proportionate either (to be parallel).

Could someone help me understand where I've gone wrong?

score 2 · Answer 1 · answered Nov 12 '22 at 09:51

2

Comment: Are not your door hinge planes perpendicular to the floor?

answered Nov 12 '22 at 09:51

Narasimham

40,495

1

Thanks. Need to learn it up now, not my cup of tea – Narasimham Nov 12 '22 at 10:32

Joshua Woo · Answer 2 · 2022-11-12T14:20:44.357

1

Two planes $z=0$ and $z=2x$ aren't mutually perpendicular or parallel, yet the plane $y=0$ is perpendicular to each of them. We don't require the three planes to be mutually perpendicular to each other, we just want one of them to be perpendicular with rest two.

For the question, one of the normal vector for both of planes is each $(2, 3, -2)$ and $(1, 2, -3)$. Say a normal vector of the plane you want to find is $(a, b, c)$. Then, we have both

$$(2, 3, -2)\cdot (a, b, c)=2a+3b-2c=0$$ $$(1, 2, -3)\cdot (a, b, c)=a+2b-3c$$ Solving the system, we get $a=5k, b=-4k, c=-k$. So, a normal vector for the plane we are looking is $(5, -4, -1)$. Any plane in the form of $5x-4y-z=m$ where $m$ is any real number is a plane that satisfies the conditions.

edited Nov 12 '22 at 14:20

answered Nov 12 '22 at 09:36

Joshua Woo

1,183

Yes. My bad. I edited it. – Joshua Woo Nov 12 '22 at 13:20
It works with cross product too : $(2,3,-2)\times(1,2,-3)=(-5,4,1)$. – Stéphane Jaouen Nov 12 '22 at 13:23
Another thing : not "the" normal vector but "a" normal vector. Right ? – Stéphane Jaouen Nov 12 '22 at 13:23
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Yep. English isn't my native language, thanks for pointing that out. And yes, the cross product seems more easier. – Joshua Woo Nov 12 '22 at 14:18

score 0 · Answer 3 · answered Nov 12 '22 at 09:47

the only condition is to impose the dot product between the normal vectors equal to $0$.
Let $v=(2,3,-2)^{T}$ the normal vector of the first plane; $w=(1,2,-3)^{T}$ the normal vector of the second plane and $t=(a,b,c)^{T}$ the normal vector of the unknown plane. We impose:
\begin{equation} \begin{cases} v \cdot t=0 \\ w \cdot t=0 \end{cases} \end{equation} \begin{equation} \begin{cases} 2a+3b-2c=0 \\ a+2b-3c=0 \end{cases} \end{equation} Then you solve the system and extract a base and you have the normal vector of the unknown plane.

score 0 · Answer 4 · answered Dec 15 '23 at 17:13

Recall the Converse of the Perpendicular Transversal Theorem for lines in 2D.

Given 2 lines are cut by a transversal, if the transversal is perpendicular to both lines, then the 2 lines must be parallel.

Now extrapolate this concept into 3D where the lines are now planes.

Given 2 planes cut by a transversal plane, if the transversal plane is mutually perpendicular to both planes, then the 2 planes must be mutually parallel.

$3$d, geometry, planes.

4 Answers4