Find theta from the cross product and dot product of two vectors

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Homework Help Overview

The discussion revolves around finding the angle theta between two vectors using their cross product and dot product. The cross product is given as 3i + j + 4k, and the dot product is 4. Participants explore the relationship between these products and the angle without deriving the original vectors.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the impossibility of retrieving the original vectors from the given products. They consider how to relate the magnitudes of the vectors to the sine and cosine of the angle theta. There is an exploration of using trigonometric identities to express tan(theta) in terms of the given values.

Discussion Status

The discussion has progressed with participants providing insights into isolating terms and relating the sine and cosine functions to the known quantities. Some participants express confusion about the steps, while others suggest methods to manipulate the equations to find tan(theta).

Contextual Notes

Participants note the challenge of working with the equations derived from the cross and dot products, emphasizing the need to compute magnitudes and the implications of treating the cross product as a numerical value rather than a vector.

loganblacke
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Homework Statement


If the cross product of vector v cross vector w = 3i + j + 4k, and the dot product of vector v dot vector w = 4, and theta is the angle between vector v and vector w, find tan(theta) and theta.


Homework Equations



vector c = |v||w| sin(theta) where vector c is the cross product of v and w.

The Attempt at a Solution



I'm assuming you have to split the cross product back into the two original vectors and then calculate the angle but I'm not sure how to go from cross product to 2 vectors. Please help!
 
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You can't get the two vectors. And you don't have to.
|3i + j + 4k|=|v|*|w|*sin(theta). 4=|v|*|w|*cos(theta). How would you get tan(theta) from that?
 
Dick said:
You can't get the two vectors. And you don't have to.
|3i + j + 4k|=|v|*|w|*sin(theta). 4=|v|*|w|*cos(theta). How would you get tan(theta) from that?

I honestly have no idea.
 
Think trig identity.
 
vela said:
Think trig identity.

That's coy. :) What's the definition of tan(theta)?
 
Dick said:
That's coy. :) What's the definition of tan(theta)?

tan theta is sin theta/cos theta.. which I think would put the vector over its magnitude and result in tan theta = unit vector..
 
loganblacke said:
tan theta is sin theta/cos theta.. which I think would put the vector over its magnitude and result in tan theta = unit vector..

? Divide the two sides of the equations by each other. Can't you find a way to get tan(theta) on one side?
 
Dick said:
? Divide the two sides of the equations by each other. Can't you find a way to get tan(theta) on one side?

I'm completely lost right now, the only thing i can work out on paper is if you isolate |v|*|w| in both equations by dividing both sides by cos theta and sin theta respectively. Then you could set the vector/sin theta = 4/cos theta.
 
loganblacke said:
I'm completely lost right now, the only thing i can work out on paper is if you isolate |v|*|w| in both equations by dividing both sides by cos theta and sin theta respectively. Then you could set the vector/sin theta = 4/cos theta.

There aren't any vectors here anymore, there's only |3i + j + 4k|. That's number, not a vector. You can compute it. Can't you get sin(theta)/cos(theta) on one side and a number on the the other?
 
Last edited:
  • #10
Dick said:
There aren't any vectors here anymore. Everything is just numbers. Sure isolate |v|*|w| in both equations. Then set the other sides equal to each other. What's the resulting equation?

I see now that its the magnitude of vector 3i + J + 4k rather than the vector itself. So you end up with sqrt(3^2+1^2+4^2)/sin theta = 4/cos theta..

So you end up with tan theta = sqrt(26)/4.
 
  • #11
then theta = arctan(sqrt(26)/4)

Thanks for the help.. again.
 
  • #12
Dick said:
That's coy. :)
I am nothing if not coy. :wink:
 

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