How to find angle between vectors from dot and cross product

In summary, to calculate the angle between two vectors when given only the value of their scalar product and the magnitude of their cross product, one must use the trigonometric identities and the fact that the tangent function is equal to the ratio of sine and cosine. After solving for the angle using the given values, it is important to consider the sign of the scalar product to determine the correct angle between the vectors.
  • #1
andylatham82
11
4
Hi, hopefully a quick question here...how do you calculate the angle between two vectors if the only information you have is the value of their scalar product and the magnitude of their cross product?

Thanks!
Andy
 
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  • #2
What do you know about the magnitude of scalar and vector product as function of the vectors and their angle?
 
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  • #3
Hi andy:
I think I remember this correctly, but you should see if you can find some verification. Wikipedia can probably help.

The magnitude of the cross product is
A × B,​
and the scalar product is
A × B × cos θ.​

Hope this helps.

Regards,
Buzz
 
  • #4
mfb said:
What do you know about the magnitude of scalar and vector product as function of the vectors and their angle?

I know that:

A.B = ABcosθ

and

|AxB| = ABsinθ

I'm just not sure where to go from there with no magnitudes or components to work with. I have a value of -7 for the scalar product (so I know the angle is greater than 90°) and a magnitude of 9 for the vector product. I'm sure there's something simple staring me in the face, but please bear with me, I'm returning to the subject of physics (and hence maths) 11 years after having last done it!
 
  • #5
You know (AB)cosθ = -7 and (AB)sinθ=9.
Can you solve for θ?

@Buzz Bloom: You forgot a sine.
 
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  • #6
I can combine the two equations to give

(sin θ)/(cos θ) = (|AxB|)/AB)

But I'm not sure where to go next. If we were talking about a right-angled triangle, we could say sin/cos = tan, but that's not the case here is it? I could do cos θ = sin (90-θ) perhaps but my algebra is failing me in reducing the result to an expression giving θ.

θ/(90-θ) = sin-1(|AxB|/AB)
 
  • #7
Hi andy:

I confess mfb did a better job than I did in advising you.

I will give you a hint similar to but a bit more direct than mfb's. You want to get a value for an expression involving
θ, but not involving either A or B. Then you solve for the angle θ.

What do you know about trigonometric identities? Wikipedia can help you.

Regards,
Buzz
 
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  • #8
andylatham82 said:
If we were talking about a right-angled triangle, we could say sin/cos = tan, but that's not the case here is it?
That is true for every angle. ##\sin##, ##\cos## and ##\tan## are just functions of an angle. It doesn't matter where the angle is.
 
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  • #9
andylatham82 said:
I can combine the two equations to give

(sin θ)/(cos θ) = (|AxB|)/AB)
Good so far.
But I'm not sure where to go next. If we were talking about a right-angled triangle, we could say sin/cos = tan, but that's not the case here is it?
You consider functions - there is no triangle involved. The tangent function is defined as this ratio.
I could do cos θ = sin (90-θ) perhaps but my algebra is failing me in reducing the result to an expression giving θ.

θ/(90-θ) = sin-1(|AxB|/AB)
That does not work. ##\sin^{-1} \left( \frac{\sin x}{\sin y} \right) \neq \frac{x}{y}##.
(if your step would work, you could multiply by the denominator to get a linear equation - but it doesn't work).
 
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  • #10
SlowThinker said:
That is true for every angle. ##\sin##, ##\cos## and ##\tan## are just functions of an angle. It doesn't matter where the angle is.

Well I finally got to the bottom of it then thanks to learning that tan = sin/cos.

sin θ/cos θ = |AxB|/A.B

⇒ tan θ = |AxB|/A.B = 9/-7

⇒ θ = 52° or 180° - 52° = 128°

Since the A.B < 0, the angle between the vectors must be greater than 90°, thus θ = 128°.

Phew! Thanks so much for your help everyone! I now know that I need to brush up on my trigonometry!
 
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1. How do I calculate the angle between two vectors using the dot product?

To find the angle between two vectors using the dot product, first calculate the dot product of the two vectors. Then, use the formula cosθ = (a · b) / (|a| * |b|) to find the cosine of the angle θ. Finally, use inverse cosine (arccos) to find the angle in radians or degrees.

2. What is the difference between the dot product and the cross product?

The dot product of two vectors results in a scalar value, while the cross product of two vectors results in a vector. The dot product measures the similarity or projection of one vector onto another, while the cross product measures the perpendicularity between two vectors.

3. Can I find the angle between two vectors using the cross product?

No, the cross product does not directly give the angle between two vectors. However, you can use the cross product and the magnitude of the two vectors to find the sine of the angle between them, and then use inverse sine (arcsin) to find the angle in radians or degrees.

4. Are there any special cases when calculating the angle between vectors using dot and cross product?

Yes, there are a few special cases to keep in mind. If the dot product is zero, the two vectors are perpendicular and the angle between them is 90 degrees. If the cross product is zero, the two vectors are parallel and the angle between them is either 0 or 180 degrees. Additionally, the order of the vectors in the cross product matters, as switching the order will result in a vector with the same magnitude but opposite direction.

5. Can I use the dot and cross product to find the angle between more than two vectors?

Yes, you can use the dot and cross product to find the angle between multiple vectors by finding the dot or cross product of every pair of vectors, and then using the appropriate formula to calculate the angle between each pair. Keep in mind that the angle between multiple vectors will be dependent on the order in which they are multiplied, so it may be helpful to use a diagram or visualize the vectors in 3D space.

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