Dot Product/Cross Product Interpretation, Geometric Construction

In summary, the homework statement asks for someone to find the vector x that is perpendicular to the cross product of a and c. However, they are not given the formula for the cross product, so they would need to use basic trig and geometry to figure it out.
  • #1
dr721
23
0

Homework Statement



Given the nonzero vector a ε ℝ3, a[itex]\dot{}[/itex]x = b ε ℝ, and a × x = c ε ℝ3, can you determine the vector x ε ℝ3? If so, give a geometric construction for x.

Homework Equations



a[itex]\dot{}[/itex]x = ||a||||x||cos[itex]\Theta[/itex]

The Attempt at a Solution



I'm not really certain what it is asking for?

Obviously, the cross product of the two vectors creates a vector perpendicular to all vectors in the a, x plane. And, the magnitude of the cross product defines the area of a parallelogram spanned by a and x.

Also, ||x||cos[itex]\Theta[/itex] is the length of the projection of x onto a, which is also equal to b/||a||

But while I know all this, I don't know what I'm trying to show or how to show it?

Any help would be great! Thanks!
 
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  • #2
Let's say you know the vector [itex]a[/itex] and all of its components. You know its dot product and cross product with [itex]x[/itex]. Can you use this information to actually figure out what each of the components of [itex]x[/itex] should be? Can you describe a picture in which it's clear how to figure out what [itex]x[/itex] should be?
 
  • #3
Muphrid said:
Let's say you know the vector [itex]a[/itex] and all of its components. You know its dot product and cross product with [itex]x[/itex]. Can you use this information to actually figure out what each of the components of [itex]x[/itex] should be? Can you describe a picture in which it's clear how to figure out what [itex]x[/itex] should be?

Well, it would be a vector perpendicular to its cross product. And, I suppose it would be a linear combination of the vector a.

I know a lot of tedious algebra might describe it, but I guess I don't really know how I would describe a picture per se. I don't know how I would break it down to each of its components.
 
  • #4
Right, it's not something that's immediately obvious in an an abstract sense, so let me rephrase the question.

If you have a point on the unit circle and you know both the sine of the angle it makes with the +x axis and the cosine as well, do you know the point's location on the unit circle absolutely?

Do you have a formula for the cross product that involves either sine or cosine?
 
  • #5
Yes, it would stand to reason that you would know that point.

And no, we have yet to learn such a formula.
 
  • #6
Hm, without knowing that [itex]|a \times x| = |a| |x| \sin \theta[/itex] where [itex]\theta[/itex] is the angle between the vectors, I think it would be difficult to make the required connection.

Let's imagine for a moment that [itex]a,x[/itex] lie in some plane. It stands to reason that [itex]c[/itex] as defined in the problem is perpendicular to this plane. Obviously you don't know [itex]x[/itex], but you should be able to find some other vector in the plane just with [itex]a[/itex] and [itex]c[/itex]. In particular, if you use a cross product to do this, you'll know that the result will be perpendicular to [itex]a[/itex] yet still in the plane [itex]a,x[/itex] span.

Once you're at this point, where you have two perpendicular vectors and [itex]x[/itex] must lie in the plane that they define, you should see that this is just a unit circle problem in some arbitrary plane. The dot product gives you the cosine, and the cross product gives you the sine.
 
  • #7
Well, that formula appears nowhere in my notes, memory, or book up to this point. So I would assume he wants us to be able to solve the problem without it. Would that be possible?

Otherwise, crossing a with c would give me a vector y that lies in the a, x span. But then how are you applying the two formulas to determine the components of x? Are you using a and y?
 
  • #8
That formula is one you should recognize as the area of the parallelogram formed by a and x.

It might help you to visualize the problem if you orient your coordinate system so that a points along the x-axis and x lies in the xy-plane.
 
  • #9
Right, I can see the formula as being true. My fear was that as he had yet to introduce it in class, my professor may frown upon its use.

That said, I'm a bit confused about this idea of creating another vector in that same plane and using it to solve for x. I don't see how the formulas allow for that.
 
  • #10
I wouldn't worry about using the formula. In fact, he might consider it something you should be able to deduce using basic trig and geometry from the fact that the magnitude of c is the area of the parallelogram.
 

1. What is the dot product?

The dot product is a mathematical operation that takes two vectors and returns a scalar quantity. It is calculated by multiplying the corresponding components of the two vectors and then adding the results together.

2. What is the cross product?

The cross product is another mathematical operation that takes two vectors and returns a third vector that is perpendicular to both of the original vectors. It is calculated by finding the determinant of a 3x3 matrix made up of the components of the two vectors.

3. How is the dot product used in geometric construction?

In geometric construction, the dot product is used to determine the angle between two vectors. This is done by taking the inverse cosine of the dot product of the two vectors divided by the product of their magnitudes.

4. How is the cross product used in geometric construction?

The cross product is used in geometric construction to find a vector that is perpendicular to two given vectors. This is useful in determining the orientation of a plane or the direction of a line in 3-dimensional space.

5. Can dot and cross products be used in 2-dimensional space?

Yes, dot and cross products can be used in 2-dimensional space as well. However, the results may not always have a physical interpretation since the concept of perpendicularity is different in 2-dimensional space compared to 3-dimensional space.

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