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I How to find angle between vectors from dot and cross product

  1. Apr 4, 2017 #1
    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?

  2. jcsd
  3. Apr 4, 2017 #2


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    What do you know about the magnitude of scalar and vector product as function of the vectors and their angle?
  4. Apr 4, 2017 #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.

  5. Apr 4, 2017 #4
    I know that:

    A.B = ABcosθ


    |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!
  6. Apr 4, 2017 #5


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    You know (AB)cosθ = -7 and (AB)sinθ=9.
    Can you solve for θ?

    @Buzz Bloom: You forgot a sine.
  7. Apr 4, 2017 #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)
  8. Apr 4, 2017 #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.

  9. Apr 4, 2017 #8
    That is true for every angle. ##\sin##, ##\cos## and ##\tan## are just functions of an angle. It doesn't matter where the angle is.
  10. Apr 4, 2017 #9


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    Good so far.
    You consider functions - there is no triangle involved. The tangent function is defined as this ratio.
    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).
  11. Apr 5, 2017 #10
    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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