Using cross product to find angle between two vectors

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Homework Help Overview

The problem involves finding the angle between two vectors, \(\vec{A}\) and \(\vec{B}\), using the cross product method. The original poster expresses frustration over obtaining an angle that does not match the expected result.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the angle using the cross product and expresses confusion over the discrepancy between their result and the expected angle. They also mention a previous similar experience.
  • Some participants suggest considering the sine function's properties and the implications of having two angles that yield the same sine value.
  • Others mention using the dot product as an alternative method to verify the angle.

Discussion Status

The discussion is ongoing, with participants exploring different methods to determine the angle and questioning the uniqueness of the angle derived from the sine function. Some guidance has been offered regarding the nature of angles and the use of the dot product.

Contextual Notes

The original poster notes a pattern in their calculations where they consistently arrive at an angle that is supplementary to the expected result, indicating a potential misunderstanding of the relationship between the angles derived from the cross product and the actual angle between the vectors.

yayscience
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Homework Statement


Find the angle between
\begin{align*}<br /> \vec{A} = 10\hat{y} + 2\hat{z} \\<br /> and \\<br /> \vec{B} = -4\hat{y}+0.5\hat{z} <br /> \end{align*}
using the cross product.

The answer is given to be 161.5 degrees.

Homework Equations


<br /> \left| \vec{A} \times \vec{B} \right| = \left| \vec{A} \right| \left| \vec{B} \right|sin(\theta)<br /> <br />

The Attempt at a Solution


<br /> \left| \vec{A} \times \vec{B} \right| = \left|<br /> \begin{array}{ccc}<br /> \hat{x} &amp; \hat{y} &amp; \hat{z} \\<br /> 0 &amp; 10 &amp; 2 \\<br /> 0 &amp; -4 &amp; 0.5<br /> \end{array} \right| = \left| 13\hat{x} \right| = 13

The magnitude of A cross B is 13.

Next we find the magnitude of vectors A and B:
\left| \vec{A} \right| = \sqrt{10^2+2^2} = \sqrt{104} = 10.198039
and
\left| \vec{B} \right| = \sqrt{(-4)^2+(\frac{1}{2})^2} = \sqrt{16.25} = 4.0311289

multiplying the previous two answers we get:
41.109609

So now we should have:
\frac{13}{41.109609} = sin(\theta)

Solving for theta, we get:
18.434951 degrees.


This is frustrating: 180-18.434951 = the correct answer. I'm not quite sure where I'm going wrong here.

I must be making the same mistake repeatedly. Another problem was the same thing, but with the numbers changed, and I also got the 180-{the answer I was getting} = {the correct answer}, but when I tried the example using the SAME methodology, I got the correct answer.

Can someone please share some relevant wisdom in my direction?
 
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sin(alpha)=sin(180-alpha) Plot the two vectors and you will see what angle they enclose.ehild
 
You might use the sign of the inner dot product to see which angle you have.
 
I can plot them, and I can see the angle, but I'm interested in calculating the angle.
When I use the dot product I get the correct result, but I cannot see where my mistake is while using the cross product.
 
There is no mistake, you get the sine of the angle, but there are two angles between 0 and pi with the same sine.

ehild
 
Oh wow; I didn't even consider that the answer wasn't unique.
Thanks!
 

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