Question

Let u = [-1,1,2] and v= [5,-3,2]. Find the equation of the plane through origin that consist of all linear combinations of u and v.

Answer 1

Start with the given equation. Plug in the given vectors and then Multiply the scalars by EVERY element in the vectors after multiplying Add the vectors by adding the corresponding components.

Answer 2

What is the given equation? ax + by+ cz = 0?

Answer 3

First take any two linear combinations of \(\vec{u}\) and \(\vec{v}\). I used \(\vec{u}+\vec{v}\) and \(\vec{u}-\vec{v}\) and the answer seems to work.
\[\vec{u}+\vec{v}=\langle4,-2,4\rangle\\\vec{u}-\vec{v}=\langle-6,4,0\rangle\]
Find the normal vector, which is given by the cross product of any two vectors in the plane. These are two such vectors, so the normal vector is
\[\vec{n}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\4&-2&4\\-6&4&0\end{vmatrix}=\langle a,b,c\rangle\]
\(a,b,c\) correspond to the coefficients in the plane equation, \(ax+by+cz=0\).

Answer 4

so would a, b, and c be equal to whatever I get for i, j, and k?

Answer 5

1u + 0v and 0u + 1v would also have worked :)

Answer 6

the ijk are what we 'solve' the psuedo determinant for. in other words, the cross product gives us:\[a\hat i+b\hat j+c\hat k\]which represents the vector (a,b,c)

Answer 7

I worked it out and solved for the vector (a,b,c) to be (-4,12,-2). Do I then put it into an equation or is -4x + 12y -2z = d my final answer?

Answer 8

not quite. we want to anchor this vector to one of the given points: (x1,y1,z1) such that

a(x-x1)+b(y-y1)+c(z-z1)=0

this will then expand to some: ax + by + cz = d

Answer 9

-(5z-6x-2y)
x y z x y
[-1,1,2] -1 1
[5,-3,2] 5 -3
+(2x+10y+3z)
+6x +2y -5z
------------
8 12 -2

4,6,-1 is what i get, but lets chk with the wolf

Answer 10

http://www.wolframalpha.com/input/?i=%5B-1%2C1%2C2%5D+cross+%5B5%2C-3%2C2%5D

8,12,-2 so a scaled version of it is 4,6,-1 is fine since its the direction that counts and not the length

Answer 11

siths cross gets us:

(-16, -24, 4), which reduces to 4,6,-1 as well when we scale it by -1/4

Answer 12

I don't understand where you got the -(5z-6x-2y) from.

Answer 13

is that from x-x1?

Answer 14

i did a quick method of a 3x3 determinant is all

Answer 15

show me how you crossed it .... since its the crossing result thats important and not the method used to cross.

Answer 16

How I crossed u and v?

Answer 17

yes

Answer 18

if you wanna learn the shortcut, its simple enough to do and easy to remember if you know the 2x2 shortcut

Answer 19

I did
i j k
-1 1 2
5 -3 2
and crossed them to get -4i + 12j - 2k

Answer 20

yeah, your end result is not correct .... which means you prolly had a simple error in the process

Answer 21

spose to get 8i not -4i

Answer 22

I just realized that as you posted. So does that give me a normal? I don't understand at all what you're telling me.

Answer 23

i j k
-1 1 2
5 -3 2

i * 1 2 = i|2-(-6)| = 8i
-3 2

Answer 24

the crossing gives you a normal vector to both u and v, yes

Answer 25

lets use the point (-1,1,2) as an anchor for the normal:

4(x+1)+6(y-1)-(z-2)=0

this is fine and dandy for a plane equation, but they tend to like to see it in its expanded form soo

4x +6y -z + (6-6+2) = 0

4x +6y -z +2 = 0 or 4x +6y -z = -2

Answer 26

hmm, typo 4(1) is not 6 lol

Answer 27

pfft, been too long. the constant will be zero since its thru the origin to start with

Answer 28

So what do you do with the -2?
I understand how you got the equation now, just I don't understand what you're supposed to do with the -2

Answer 29

does it just disappear?

Answer 30

see it for the typo that it is and correct it :)

Answer 31

Oh okay! I see it now

Answer 32

4(x+1)+6(y-1)-(z-2)=0

4x+4+6y-6-z+2=0 :) 4-6+2 = 0 is your constant, not -2

Answer 33

now if we hadnt reduced the magnitude of the normal we would have gotten

8x + 12y -2z = 0 which is not in 'simplest' form since its just a multiple of what we have.

Answer 34

but either answer will work correct?

Answer 35

depends on whos grading it and what they expect from you.

Answer 36

any positive, nonzero multiple of (4x+6y-z=0) will be the equation of the plane, but its up to your own material how specific it needs to get

Answer 37

I don't imagine they'll mark it wrong if it's in its simplest form.
I have this question solved but I have a few others that I might just need a nudge in the right direction for, if I need help I will post it

Answer 38

very well :)