Creating a question with vectors in 3-space (perpendicular)

[Physics345]

Homework Statement

Describe how the dot product can be used to determine whether two vectors are perpendicular. Create a question with vectors in 3-space to illustrate this property. Solve the question as well.

Homework Equations

The Attempt at a Solution

Did I do this right? I did something I've done/never seen before in my text, basically I used the cross product to find a final perpendicular vector, does this work?

 

[tnich]

Homework Statement

Describe how the dot product can be used to determine whether two vectors are perpendicular. Create a question with vectors in 3-space to illustrate this property. Solve the question as well.

Homework Equations

The Attempt at a Solution

Did I do this right? I did something I've done/never seen before in my text, basically I used the cross product to find a final perpendicular vector, does this work?

View attachment 222570

You have done a great job on your question and your solution.

 

[tnich]

You have done a great job on your question and your solution.

I think you could spruce up the problem statement a little. For example, you could say, given two perpendicular vectors $\vec u = (1, 2, 1)$ and $\vec v = (-2, 2, -2)$, find a third vector $\vec w$ that is perpendicular to both $\vec u$ and $\vec v$ and show that all three pairs of vectors are perpendicular.

 

[Physics345]

You have done a great job on your question and your solution.

Eureka it does work! good to know. Oh how I love math, it's so satisfying learning math! Thank you!

 

[Physics345]

I think you could spruce up the problem statement a little. For example, you could say, given two perpendicular vectors $\vec u = (1, 2, 1)$ and $\vec v = (-2, 2, -2)$, find a third vector $\vec w$ that is perpendicular to both $\vec u$ and $\vec v$ and show that all three pairs of vectors are perpendicular.

I like! That sounds like fun. I'll give it a go once I've completed my reply on another thread.

 

[Physics345]

I think you could spruce up the problem statement a little. For example, you could say, given two perpendicular vectors $\vec u = (1, 2, 1)$ and $\vec v = (-2, 2, -2)$, find a third vector $\vec w$ that is perpendicular to both $\vec u$ and $\vec v$ and show that all three pairs of vectors are perpendicular.

Now there is an issue, I can't think of another way to write that other than the way you have.

 

[Physics345]

I got it!
Given two vectors find a third perpendicular vector and then prove that all three pairs are perpendicular to each other.
Even though it is pretty much the same thing, it is more or less in my words.

 

[Ray Vickson]

Homework Statement

Describe how the dot product can be used to determine whether two vectors are perpendicular. Create a question with vectors in 3-space to illustrate this property. Solve the question as well.

Homework Equations

The Attempt at a Solution

Did I do this right? I did something I've done/never seen before in my text, basically I used the cross product to find a final perpendicular vector, does this work?

View attachment 222570

You have not solved the problem that was ASKED: it wanted you to use the DOT product, not the cross product!

Besides, you can use the dot product involving two vectors to construct a new vector perpendicular to them both in any dimension ≥ 3, but most such higher-dimensional spaces do not allow anything like a cross-product. (If I remember correctly, you can also have a cross-product in 7 dimensions, but not in any other!)

 

[tnich]

I got it!
Given two vectors find a third perpendicular vector and then prove that all three pairs are perpendicular to each other.
Even though it is pretty much the same thing, it is more or less in my words.

I think you need to specify that the two original vectors are perpendicular 3-D vectors for you method to work.

 

[Physics345]

You have not solved the problem that was ASKED: it wanted you to use the DOT product, not the cross product!

Besides, you can use the dot product involving two vectors to construct a new vector perpendicular to them both in any dimension ≥ 3, but most such higher-dimensional spaces do not allow anything like a cross-product. (If I remember correctly, you can also have a cross-product in 7 dimensions, but not in any other!)

I just used the cross product to find a third vector and then found out if they are perpendicular with the dot product.

 

[Physics345]

I think you need to specify that the two original vectors are perpendicular 3-D vectors for you method to work.

Ill be sure to address that, thanks for pointing it out.

 

[Ray Vickson]

I just used the cross product to find a third vector and then found out if they are perpendicular with the dot product.

Yes, I know: I read your solution. However, I still maintain that you would learn more by trying to find a vector perpendicular to u and v but without using the cross product. We sometimes need to do this type of thing in spaces of dimension 15 or 20 or more, and so it is important to know ho to do it.

 

[Physics345]

Yes, I know: I read your solution. However, I still maintain that you would learn more by trying to find a vector perpendicular to u and v but without using the cross product. We sometimes need to do this type of thing in spaces of dimension 15 or 20 or more, and so it is important to know ho to do it.

I'm open to giving it a try, would you mind guiding me through the process?

 

[Mark44]

I'm open to giving it a try, would you mind guiding me through the process?

The vectors in your example are <1, 2, 1> and <-2, 2, -2>.
Let the vector to be found be u = <a, b, c>
Dot each of the two given vectors with u. This will give you two equations in three unknowns. There will be an infinite number of solutions, because all you need is a vector that is perpendicular to the two given vectors, so any scalar multiple of the vector you find will also be perp. to the two given vectors.

 

[Ray Vickson]

I'm open to giving it a try, would you mind guiding me through the process?

Google "Gram-Schmidt" or "orthogonal projection".