My Math Forum Simple vector angle question gone wrong

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 March 6th, 2016, 08:17 AM #1 Senior Member   Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 Simple vector angle question gone wrong Hello, I don't even know where to put this question because it's quite basic, but I get a different answer than the one in the book: I have 2 vectors, AM and AN: AM=w+0.5u AN=w+0.5v Also, |u|=|v|=|w|=1. What is the angle MAN? I did (AMAN)/(|AM|*|AN|) and got that cos(MAN)=0.8, thus MAN=36.86, but the angle in the book is 25.84, which fits to cos(MAN)=0.9, where is my mistake? Thank you!
 March 6th, 2016, 08:29 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,723 Thanks: 1376 I agree with your calculation ... maybe a book typo Thanks from noobinmath
 March 6th, 2016, 08:46 AM #3 Senior Member   Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 Thank you sir
 March 6th, 2016, 11:52 AM #4 Newbie   Joined: Jan 2016 From: Ghana Posts: 17 Thanks: 4 Could one of you show your working on how you got the angle? I did it and got a different angle, I think I may have missed a step or something
 March 6th, 2016, 12:06 PM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 2,723 Thanks: 1376 $\vec{AM} = 0.5u + 0v + w$ $\vec{AN} = 0u + 0.5v + w$ $\vec{AM} \cdot \vec{AN} = (0.5)(0) + (0)(0.5) + (1)(1) = 1$ $|AM| = \sqrt{0.5^2 + 0^2 + 1^2} = \sqrt{1.25}$ $|AN| = \sqrt{0^2 + 0.5^2 + 1^2} = \sqrt{1.25}$ $\cos{\theta} = \dfrac{\vec{AM} \cdot \vec{AN}}{|AM||AN|} = \dfrac{1}{1.25} = 0.8$ $\theta = \arccos(0.8 ) \approx 37^\circ$ Thanks from noobinmath and Lanthanide

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