Magnitude and direction of a resultant vector.

Problem: Find the magnitude and direction of the resultant of two vectors $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ with magnitudes of 6 and 8 and an angle of 60° between them.

Okay, so the first thing I do is find θ by subtracting 60° from 180° and yield 120°. Then I use cosine law to find the magnitude:

$\displaystyle | \vec{u}+\vec{v} |^{2} = | \vec{u} |{^2}+| \vec{v} |^{2} - 2 | \vec{u} | | \vec{v} | cos\theta$

$\displaystyle | \vec{u}+\vec{v} |^{2} = (6)^{2} + (8)^{2} - 2|6||8|cos(120°)$

$\displaystyle | \vec{u}+\vec{v} |^{2} = \sqrt{148}$

Alright, so I got the magnitude right according to my book, and then I go on to draw the vectors, tail-to-tail and then the resultant: Vector diagram

So, now I use sine law to find angle alpha:

$\displaystyle \frac{sin\alpha}{|\vec{u}|} = \frac{sin120°}{|\vec{u}+\vec{v}|} $

$\displaystyle \frac{\sin\alpha }{6} = \frac{\sin120°}{\sqrt{148}}$

$\displaystyle \sin\alpha =\frac{(6)\sin120°}{\sqrt{148}} $

$\displaystyle \sin\alpha \approx 0.4271$

$\displaystyle \alpha \approx \sin^{-1}(0.4271) \approx 25.285$° is the direction I get after using sine law, but the answer is completely different in my textbook. The reason why is because they drew $\displaystyle \vec{u}$ first then $\displaystyle \vec{v}$ tail to tail, but I did not learn any law in which there is a general convention to draw vectors of differing magnitudes from tail to tail, therefore I dismissed it as arbitrary. If there is no law, then by using sine law, shouldn't I have arrived at the same direction as the book? The answer in the book:

α ≈ 34.7°, "the resultant of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ is approximately 34.7° from $\displaystyle \vec{u}$."

So, where did I go wrong? Thanks in advance.

http://imageshack.us/photo/my-images...eanddirec.jpg/