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Dot Product/Cross Product Interpretation, Geometric Construction

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dr721
#1
Sep4-12, 11:16 PM
P: 23
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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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Muphrid
#2
Sep4-12, 11:36 PM
P: 834
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?
dr721
#3
Sep4-12, 11:48 PM
P: 23
Quote Quote by Muphrid View Post
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.

Muphrid
#4
Sep4-12, 11:54 PM
P: 834
Dot Product/Cross Product Interpretation, Geometric Construction

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?
dr721
#5
Sep5-12, 12:03 AM
P: 23
Yes, it would stand to reason that you would know that point.

And no, we have yet to learn such a formula.
Muphrid
#6
Sep5-12, 12:12 AM
P: 834
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.
dr721
#7
Sep5-12, 12:27 AM
P: 23
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?
vela
#8
Sep5-12, 06:02 AM
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PF Gold
P: 11,722
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.
dr721
#9
Sep5-12, 07:20 AM
P: 23
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.
vela
#10
Sep5-12, 12:34 PM
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PF Gold
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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.


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