# Dot Product/Cross Product Interpretation, Geometric Construction

by dr721
Tags: construction, geometric, interpretation, product, product or cross
 P: 23 1. The problem statement, all variables and given/known data 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. 2. Relevant equations a$\dot{}$x = ||a||||x||cos$\Theta$ 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$\Theta$ 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!
 P: 834 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?
P: 23
 Quote by 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?
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.

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?
 P: 23 Yes, it would stand to reason that you would know that point. And no, we have yet to learn such a formula.
 P: 834 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.
 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?
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,923 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.
 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.
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,923 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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