Dot Product/Cross Product Interpretation, Geometric Construction

dr721

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!

Muphrid

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

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

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

Yes, it would stand to reason that you would know that point.

And no, we have yet to learn such a formula.

Muphrid

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

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

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

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

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.