# Cartesian, vector equations

#### Kaspelek

##### New member
-

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.

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#### MarkFL

Staff member
Hello and welcome to MHB, Kaspelek! 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

##### New member
Hello and welcome to MHB, Kaspelek! 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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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

##### New member
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Thanks for the quick response,

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

##### New member
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)

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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

##### New member
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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

##### New member
Thanks so much for your assistance