Determining 3D Vector Basis with a,b,c Vectors

In summary, to determine if the vectors a = (2, -3, 2), b = (1, 1, -1), and c = (8, 5, -2) can be used as a basis for vectors in R^3, one can use the dot product to check if they are mutually orthogonal. However, they do not have to be orthogonal to be linearly independent. The scalar triple product can also be used to determine if the vectors are linearly independent. A basis for R^3 consists of three linearly independent vectors.
  • #1
kevykevy
25
0

Homework Statement


Determine whether the the vectors a = (2, -3,2), b = (1, 1, -1) and
c = (8, 5, -2) can be used as a basis for vectors in R^3 (3D space)


Homework Equations





The Attempt at a Solution


I really have no clue, I think maybe you use either cross product, dot product or triple scalar product...?
 
Physics news on Phys.org
  • #2
kevykevy said:
I think maybe you use either cross product, dot product or triple scalar product...?

Why don't you try one of these? I'd use the dot product first, to show whether or not the vectors are mutually orthogonal.
 
  • #3
kevykevy said:

Homework Statement


Determine whether the the vectors a = (2, -3,2), b = (1, 1, -1) and
c = (8, 5, -2) can be used as a basis for vectors in R^3 (3D space)

What's the definition of a basis?
 
  • #4
cristo said:
Why don't you try one of these? I'd use the dot product first, to show whether or not the vectors are mutually orthogonal.

They don't have to be mutually orthogonal to be linearly independant, and it is unlikely that they will be. To kevykevy: You were on the right track with the scalar triple product. What properties of this product do you know?
 
  • #5
LeonhardEuler said:
They don't have to be mutually orthogonal to be linearly independant, and it is unlikely that they will be. To kevykevy: You were on the right track with the scalar triple product. What properties of this product do you know?

Sorry, I read "orthogonal" that wasn't in the question!
 
  • #6
cristo said:
Sorry, I read "orthogonal" that wasn't in the question!

I know what you're talking about. I've been there more than a few times myself. :redface:
 
  • #7
Cross Product
a x b = (1, 4, 5)

Dot Product
(1, 4, 5) x (8, 5, -2) = 18

Since 18 doesn't equal 0, then the vectors cannot be used as basis vectors

is that right?
 
  • #8
to radou - basis vectors, example i, j, and k with the carot(^) on top
 
  • #9
kevykevy said:
to radou - basis vectors, example i, j, and k with the carot(^) on top

Ok, that's an example of a basis. We can add that every set consisting of three linearly independent vectors forms a basis for R^3. All you have to do is check if your vectors are linearly independent.
 
  • #10
And if you are going to be doing problems like this it would be a really good idea for you to look at the definition of "basis" in your textbook.
 

1. What is a vector basis?

A vector basis is a set of linearly independent vectors that can be used to represent any vector in a given vector space.

2. How do you determine a 3D vector basis?

To determine a 3D vector basis, you need to have three linearly independent vectors in three-dimensional space. These vectors can be represented as a,b, and c. Then, you can use the Gram-Schmidt process or other methods to find the orthonormal basis vectors that span the same subspace as a,b, and c.

3. What is the Gram-Schmidt process?

The Gram-Schmidt process is a method for finding an orthonormal basis from a set of linearly independent vectors. It involves a series of orthogonalization and normalization steps to create a set of vectors that are perpendicular to each other and have a length of 1.

4. Why is it important to have an orthonormal basis?

An orthonormal basis is important because it simplifies vector calculations and makes them more intuitive. It also allows for easier visualization and interpretation of vector operations.

5. Can a set of vectors that are not linearly independent be used as a basis?

No, a set of vectors that are not linearly independent cannot be used as a basis. This is because they do not span the entire vector space and, therefore, cannot be used to represent all possible vectors in that space.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
5
Views
460
  • Precalculus Mathematics Homework Help
Replies
21
Views
7K
  • Precalculus Mathematics Homework Help
Replies
18
Views
461
  • Precalculus Mathematics Homework Help
Replies
15
Views
2K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
2K
  • Precalculus Mathematics Homework Help
Replies
7
Views
2K
  • Precalculus Mathematics Homework Help
Replies
14
Views
5K
  • Precalculus Mathematics Homework Help
Replies
16
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
3K
Back
Top