# Thread: finding angle using cross product and dot product

1. ## finding angle using cross product and dot product

If = 3 - 4 + 6, and · = 4, find tanθ where θ is the angle between and . Give an exact answer.

not sure what to do here, i know that i can divide sin by cos to get tan but i get a vector... help please

2. Originally Posted by acosta0809
If = 3 - 4 + 6, and · = 4, find tanθ where θ is the angle between and . Give an exact answer.

not sure what to do here, i know that i can divide sin by cos to get tan but i get a vector... help please
$|\vec{V}| |\vec{W}| \sin \theta = |\vec{V} \times \vec{W} | = \sqrt{3^2+(-4)^2+6^2}=\sqrt{61}$
$|\vec{V}| |\vec{W}| \cos \theta = \vec{V} \cdot \vec{W} = 4$
$\tan \theta=\frac{\sqrt{61}}{4}$

3. when you have the abs of v cross w on the top one why is it absolute value i thought that v cross w is equal

$|\vec{V}| |\vec{W}| \sin \theta$

4. Originally Posted by acosta0809

when you have the abs of v cross w on the top one why is it absolute value i thought that v cross w is equal

$|\vec{V}| |\vec{W}| \sin \theta$
First, it is NOT "absolute value" it is the length of the vector. And what you thought is wrong. $\vec{u} \times \vec{v}$ is a vector, not a number. $|\vec{v}||\vec{w}|sin(\theta)$ is the length of $\vec{u} \times \vec{v}$, NOT $\vec{u} \times \vec{v}$ itself.

5. Evaluating the dot and cross product is probably the best way to work out angles between vectors because you don't have to normalise anything. But the tangent function has a period of pi radians, so when you divide to work out the tangent of the angle you are losing some information. Note that the signs of the dot and cross product together tells you which quadrant the angle comes from. So in fact you can work out the angle without any ambiguity at all. Math libraries for programming languages like C/C++ usually have an atan2() function which is just what you want to get the angle properly.

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# how to find theta using dot product

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