Using cross product to find angle between two vectors

In summary, the angle between vector A and B using the cross product is not unique and can have two possible values. The correct value can be found by taking the inverse sine of the ratio of the magnitude of the cross product of A and B to the product of the magnitudes of A and B.
  • #1
yayscience
5
0

Homework Statement


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

The answer is given to be 161.5 degrees.

Homework Equations


[tex]
\left| \vec{A} \times \vec{B} \right| = \left| \vec{A} \right| \left| \vec{B} \right|sin(\theta)

[/tex]

The Attempt at a Solution


[tex]
\left| \vec{A} \times \vec{B} \right| = [/tex] [tex]\left|
\begin{array}{ccc}
\hat{x} & \hat{y} & \hat{z} \\
0 & 10 & 2 \\
0 & -4 & 0.5
\end{array} \right| = \left| 13\hat{x} \right| = 13 [/tex]

The magnitude of A cross B is 13.

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

multiplying the previous two answers we get:
41.109609

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

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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  • #2
sin(alpha)=sin(180-alpha) Plot the two vectors and you will see what angle they enclose.ehild
 
  • #3
You might use the sign of the inner dot product to see which angle you have.
 
  • #4
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.
 
  • #5
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
 
  • #6
Oh wow; I didn't even consider that the answer wasn't unique.
Thanks!
 

FAQ: Using cross product to find angle between two vectors

What is a cross product?

A cross product is a mathematical operation that takes two vectors as input and produces a new vector that is perpendicular to both of the input vectors.

How is cross product used to find the angle between two vectors?

The magnitude of the cross product of two vectors is equal to the product of the magnitudes of the two vectors multiplied by the sine of the angle between them. By rearranging this formula, we can solve for the angle between the two vectors.

Can cross product be used to find the angle between any two vectors?

No, cross product can only be used to find the angle between two vectors in three-dimensional space. In two-dimensional space, the concept of cross product does not exist.

What is the difference between cross product and dot product?

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

Is cross product commutative?

No, cross product is not commutative. This means that the order in which the two vectors are multiplied matters and will produce different results.

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