Dot Product/Cross Product Interpretation, Geometric Construction

[dr721]

Homework Statement

Given the nonzero vector a ε ℝ³, a\dot{}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\dot{}x = ||a||||x||cos\Theta

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\Theta 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 a and all of its components. You know its dot product and cross product with x. Can you use this information to actually figure out what each of the components of x should be? Can you describe a picture in which it's clear how to figure out what x should be?

 

[dr721]

Let's say you know the vector a and all of its components. You know its dot product and cross product with x. Can you use this information to actually figure out what each of the components of x should be? Can you describe a picture in which it's clear how to figure out what x 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]

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 |a \times x| = |a| |x| \sin \theta where \theta is the angle between the vectors, I think it would be difficult to make the required connection.

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

Once you're at this point, where you have two perpendicular vectors and x 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.