How Do You Convert a Cartesian Equation to a Vector Equation in R3?

[Kaspelek]

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Hi guys,

I'm new to this forum and was just wondering if I could receive some help on this question.

I'm really struggling to complete it.

Consider the plane in R3 with the Cartesian equation
x + 7y − 2z = 0.
(You may assume that this is a subspace of R3.)
(a) Find a vector equation for the plane.
(b) Use (a) to find a finite spanning set for the plane.

Thanks in advance guys!

 

[MarkFL]

Hello and welcome to MHB, Kaspelek! :D

Can you show us what you have done so far? This way our helpers know where you are stuck and can offer specific help.

 

[Kaspelek]

Hello and welcome to MHB, Kaspelek! :D

Can you show us what you have done so far? This way our helpers know where you are stuck and can offer specific help.

So i found 3 points that exist on that plane

A(2,2,8) B(4,2,9) C(8,2,1)

u= B-A =(2,0,1)
v= C-A= (6,0,3)

Therefore vector equation r=(2,2,8)+s(2,0,1)+ t(6,0,3)

If that is a correct vector equation, I'm not sure how to do part b.

 

[I like Serena]

So i found 3 points that exist on that plane

A(2,2,8) B(4,2,9) C(8,2,1)

u= B-A =(2,0,1)
v= C-A= (6,0,3)

Therefore vector equation r=(2,2,8)+s(2,0,1)+ t(6,0,3)

Good!
However, your vectors within the plane need to be independent.
In your case (6,0,3)=3*(2,0,1), meaning they are not independent.
Can you find another vector within the plane that is independent?

If that is a correct vector equation, I'm not sure how to do part b.

Part (b) is just about definitions.
The 2 vectors you would have for (a) "span" the plane.

 

[Kaspelek]

Good!
However, your vectors within the plane need to be independent.
In your case (6,0,3)=3*(2,0,1), meaning they are not independent.
Can you find another vector within the plane that is independent?
Part (b) is just about definitions.
The 2 vectors you would have for (a) "span" the plane.

Thanks for the quick response,

What's the best way about finding 3 independent vectors in this case?

 

[I like Serena]

Thanks for the quick response,

What's the best way about finding 3 independent vectors in this case?

You already have them.
Note that (0,0,0) is also a vector in the plane.
In other words, each of your vectors to A, B, and C are vectors within the plane.
Just pick one of those vectors that is not a multiple of (2,0,1).

More generally, from your equation x+7y-2z=0, you can deduce that (1,7,-2) is a so called normal vector.
You need 2 vectors that are perpendicular to this vector.
That means their dot product is zero.
The vector you already found is (2,0,1). Taking the dot product with (1,7,-2) is indeed zero.
Another vector might be (7,-1,0), which also has a dot product with (1,7,-2) that is zero.

 

[Kaspelek]

I found another vector B(1,5/7,3)

Therfore u=B-A=(1,5/7,3)-(2,2,8)=(-1,-9/7,-5)

Hence vector equation=r=(2,2,8)+s(-1,-9/7,-5)+t(6,0,3)

Does A need to be an independent vector too? Cause 2*(1,1,4)=(2,2,8)

 

[I like Serena]

I found another vector B(1,5/7,3)

Therfore u=B-A=(1,5/7,3)-(2,2,8)=(-1,-9/7,-5)

Hence vector equation=r=(2,2,8)+s(-1,-9/7,-5)+t(6,0,3)

Good! :)

Does A need to be an independent vector too? Cause 2*(1,1,4)=(2,2,8)

No.
This is just any vector to the plane.
Usually a multiple of this vector will not be in the plane, although in this particular case it will be.

 

[Kaspelek]

Good! :)
No.
This is just any vector to the plane.
Usually a multiple of this vector will not be in the plane, although in this particular case it will be.

So how exactly would i answer b, in the case of the above vector equation?

 

[I like Serena]

So how exactly would i answer b, in the case of the above vector equation?

The set {(-1,-9/7,-5), (6,0,3)} is a finite spanning set for the plane.

 

[Kaspelek]

Thanks so much for your assistance