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Using cross product to find angle between two vectors 
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#1
Jun3011, 12:33 AM

P: 5

1. The problem statement, all variables and given/known data
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. 2. Relevant equations [tex] \left \vec{A} \times \vec{B} \right = \left \vec{A} \right \left \vec{B} \rightsin(\theta) [/tex] 3. 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: 18018.434951 = 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? 


#2
Jun3011, 01:21 AM

HW Helper
Thanks
P: 10,776

sin(alpha)=sin(180alpha) Plot the two vectors and you will see what angle they enclose.
ehild 


#3
Jun3011, 01:57 AM

HW Helper
P: 6,189

You might use the sign of the inner dot product to see which angle you have.



#4
Jun3011, 03:05 AM

P: 5

Using cross product to find angle between two vectors
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
Jun3011, 03:19 AM

HW Helper
Thanks
P: 10,776

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
Jul111, 12:30 AM

P: 5

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


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