Vector question : acute angle? Watch

hamstein
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Hey I managed to find the magnitude of the equations to be
1. root 104
2. root 14

Is there a diagram I have to draw to find the acute angle? :confused:
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mathslover786
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a.b....
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jonny23563
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Do you know the full definition of the dot product?
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hamstein
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(Original post by mathslover786)
a.b....
Okies thanks I didn't realise you had to use that formula hah
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hamstein
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(Original post by jonny23563)
Do you know the full definition of the dot product?
Erm, finding an angle? I know the formula :confused:
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jonny23563
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(Original post by hamstein)
Erm, finding an angle? I know the formula :confused:
 \displaystyle \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|cos\theta

I know the vectors should be bold and not arrowed but I think it makes it clearer!
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hamstein
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(Original post by jonny23563)
 \displaystyle \vec{a} \cdot \vec{b} = \frac{|\vec{a}||\vec{b}|}{cos\th  eta}

I know the vectors should be bold and not arrowed but I think it makes it clearer!
Oh thanks Do you always have to used the dot formula to find an acute angle? I've seen somewhere you sometimes can use tan
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Lou Reed
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(Original post by jonny23563)
 \displaystyle \vec{a} \cdot \vec{b} = \frac{|\vec{a}||\vec{b}|}{cos\th  eta}

I know the vectors should be bold and not arrowed but I think it makes it clearer!
its a\cdot b = |a||b|\cos\theta
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jonny23563
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(Original post by hamstein)
Oh thanks Do you always have to used the dot formula to find an acute angle? I've seen somewhere you sometimes can use tan
I always use it just because it's the easiest way - you'll have a and b and it's just plugging numbers in. I've never heard of a way of using tan
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ogloom
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Thats not the formula!

its a.b=|a|*|b|*cos(theta).
If by chance you get an obtuse angle. Simply subtract it from 180. I would draw the diagram, but I'm sure you can work out why.
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jonny23563
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(Original post by Lou Reed)
its a\cdot b = |a||b|\cos\theta
nice spot :p:
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Jooeee
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(Original post by ogloom)
If by chance you get an obtuse angle. Simply subtract it from 180.
Erm no?
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hamstein
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(Original post by ogloom)
Thats not the formula!

its a.b=|a|*|b|*cos(theta).
If by chance you get an obtuse angle. Simply subtract it from 180. I would draw the diagram, but I'm sure you can work out why.
Do you use CAST? Like in trigonometry?
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ogloom
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(Original post by Jooeee)
Erm no?
Why not?
http://yfrog.com/0ztrigj. Using the dot product always gives us angle a (i.e. both vectors moving out/in). If we want to find the acute angle, simply subtract the value from 180 to find the other angle, b
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Jooeee
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(Original post by hamstein)
Do you use CAST? Like in trigonometry?
I'm pretty sure you won't be needing to do anything like that. If you need the angle, then when you do inverse of it, if you use a calculator, it will return the principal? value which is <360

The only thing you could do is that it doesn't matter if you have two vectors if you take the reflex angle, or the "smaller" angle. Since  \cos \theta=\cos(360-\theta)
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hamstein
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(Original post by ogloom)
Why not?
http://yfrog.com/0ztrigj. Using the dot product always gives us angle a (i.e. both vectors moving out/in). If we want to find the acute angle, simply subtract the value from 180 to find the other angle, b
Thanks okies so everytime I see obtuse I subtract from 180 right?
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ogloom
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Yes. Jooeee, we've been taught C4 very well. I can guarantee you that vectors don't have to be tail to tail when working out the angle.
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hamstein
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(Original post by ogloom)
Yes. Jooeee, we've been taught C4 very well. I can guarantee you that vectors don't have to be tail to tail when working out the angle.
Yes how did you get so good at C4?
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Jooeee
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(Original post by ogloom)
Yes. Jooeee, we've been taught C4 very well. I can guarantee you that vectors don't have to be tail to tail when working out the angle.
Yes ok, sorry. I was being stupid.
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hamstein
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Is there always just one angle?
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