Question

Find the angle between a and b, if the dot product is sqrt(3) and a x b=<1,2,2>

Answer 1

Ok so..
\[a \ \dot\ b=\sqrt{3} = |a||b|\cos \theta \\ |a||b|= \frac{ \sqrt{3}}{\cos \theta} \\ a X b=<1,2,2> = |a||b|\sin \theta\\ <1,2,2>* \frac{ \cos \theta}{ \sqrt{3}} =\sin \theta\]
\[\tan \theta= \frac{ <1,2,2>}{ \sqrt{3} }\] I'm not sure if this is right or what I do from here!

Answer 2

@mahmit2012 @myko @phi @turingtest

Answer 3

Almost there:
tan theta = ||<1,2,2>|| / sqrt(3)

Answer 4

Thanks!

Answer 5

You're welcome ! :)

Answer 6

a slight correction. you wrote
\[ a \times b = |a| |b| sin \theta \]
it should read:
\[ | a \times b| = |a| |b| sin \theta \]
the magnitude of the cross product. (the cross product produces a vector, and this equation tells us the magnitude of that vector)

it follows that
\[ \frac{ | a \times b|}{a \cdot b}= \frac{|a| |b| \sin\theta}{|a||b| \cos\theta}= \tan \theta \]