Question

Vector help: How do I find an equation of a plane when I am only given a point on the plane and the equations of two orthrogonal planes?
Find an equation of the plane passing though (0, −2, 4) and perpendicular to the planes
2x + 5y − 3z = 0 and x + y − 2z + 8 = 0.
(Again, guide me along the way, don't give me the direct answer)

Answer 1

Vectors orthogonal to each other have a dot product of zero, right? So I suppose we have to dot some products?

Answer 2

calc 3... I took that last semester... but I left my notes in my dorm >,<

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @HexiFlexiDexi
Vectors orthogonal to each other have a dot product of zero, right? So I suppose we have to dot some products?
\(\color{#0cbb34}{\text{End of Quote}}\)

I think so, but I would ask someone who knows this better lol

Answer 4

"Vectors orthogonal to each other have a dot product of zero"

just googled, that's right ^_^

Answer 5

Okai, thanks~ Do you guys know what to do next though?

Answer 6

absolutely no clue!!!!!! but @mathmale might have a clue xD

Answer 7

I was pretty sure that the normal vector of 2x + 5y − 3z = 0 would be <2, 5, -3> ?

Answer 8

and the normal vector of x + y − 2z + 8 = 0 would be <1, 1, -2>

do you get how I'm getting that?

Answer 9

No...?

Answer 10

when the equation of a plane is ax + by + cz = # form
the "normal vector" of the plane is just < a , b , c >

Answer 11

by normal I mean orthogonal or perpendicular ^_^"

Answer 12

Okai :)

Answer 13

no @ILovePuppiesLol most people only have to learn up to Calculus 1 or 2
this is Calculus 3

Answer 14

although I'm honestly confused by the fact that they seemed to have given two orthogonal planes >.>

Answer 15

XD I'm still in middle school. just learning ahead in a accelerated course :)

Answer 16

O_O
'accelerated' ?!
'middle school' ?!
WHAT?!
and I thought I was learning this early in my senior year of high school ;-;

Answer 17

This is Calculus? It looks like stuff I learned in high school last year.

Answer 18

repeatativity tells me to take the cross product of the two normal vectors I found above and use that as the normal vector for the equation we are looking for

Answer 19

Wait, what? There's only two equations for planes there, how do I get the vectors? By getting three points by subbing numbers as x, y, and z?

Answer 20

normal vector of 2x + 5y − 3z = 0 would be <2, 5, -3>

Answer 21

normal vector of x + y − 2z + 8 = 0 would be <1, 1, -2>

Answer 22

so we have two vectors

Answer 23

Oh, right hahaha forgot that fact :P

Answer 24

Lemme quickly cross these two vectors on "paper" :)

Answer 25

Oh, and why does the vector normal to the plane just that? (like, just take x, y, and z from it)? Can anyone geometrically explain it pls? I gtg~

Answer 26

Thanks, I'll come back to it later <3

Answer 27

\(\color{#0cbb34}{\text{Originally Posted by}}\) @HexiFlexiDexi
Oh, and why does the vector normal to the plane just that? (like, just take x, y, and z from it)? Can anyone geometrically explain it pls? I gtg~
\(\color{#0cbb34}{\text{End of Quote}}\)

I'm sure there is a proof somewhere, but I've never seen it... if I find it, I'll try to show it to you ^_^
here's an example for now: http://math.stackexchange.com/questions/352134/finding-the-vector-perpendicular-to-the-plane

Answer 28

For when you come back (I might not be on)

Take the cross product (you will get a vector)

plug it into the equation of a plane: \[<a,b,c>•(<x,y,z>-<x_0,y_0,z_0>) = 0 \]\[\rightarrow ~~<a,b,c>•<x-x_0,y-y_0,z-z_0> = 0 \]\[\rightarrow ~~a(x-x_0)+b(y-y_0)+c(z-z_0)=0\]
use the last one
plug in the normal vector that you got from the cross product as \(a,~b,\) and \(c\)
and plug in the point they gave you as \(x_0,~y_0,\) and \(z_0\)

Answer 29

@sleepyhead314 Okay, I get why the normal vector is the coefficients of the plane~
Also, uh, what was the cross product for? The dot product is to make the thingy perpendicular to the other vector, right?

Answer 30

@Astrophysics @hartnn can u guys help this user with question :)

Answer 31

@sleepyhead314 "and the normal vector of x + y − 2z + 8 = 0 would be <1, 1, -2>"
What about the 8? Or does it not matter since it is a constant?

Answer 32

Hey, can anyone here help? I'm really stuck here~

Answer 33

do you remember on one of you earlier questions, i said that if you observe closely, you will find a neat little shortcut...

Answer 34

Yea~ The constant stays the same-ohhh XD

Answer 35

i have commented again on that question, have a look :)

Answer 36

The cross product of the two given orthogonal vectors
is orthogonal to the plane you want.

Answer 37

Lets suppose the x y plane is the plane you want.
then the xz plane and yz plane orthogonal to it

Answer 38

Okai, got it~ We get a vector normal to the two given planes by the cross product, right? What is that for?

Answer 39

you are given the equations of two planes.
I am going to draw them

Answer 40

vector AB is normal to the xz plane
vector AC is normal to the yz plane.
the cross product of these two vectors , is normal to the plane we want

Answer 41

We can write the equation of a plane if we know the normal vector to it
and a point on the plane