What is the angle between <3 , -2 , 7 >  and  < 5, -3 , 2 > ?

Apr 22, 2017

the angle is approximately 43.86 degrees

Explanation:

The formula for finding the angle between vectors is:

$\vec{v} \cdot \vec{u} = | \vec{v} | \times | \vec{u} | \cos \theta$, where $\vec{v} \cdot \vec{u}$ is the dot product of the two vectors, and $| \vec{v} |$ and $| \vec{u} |$ are the magnitudes of the two vectors.

So first we need to find the dot product of the two vectors.

The dot product can be found simply adding up each corresponding element: $3 \times 5 + \left(- 2\right) \times \left(- 3\right) + 7 \times 2 = 15 + 6 + 14 = 35$

The magnitude can be found by square rooting the sum of the squares of the elements of a vector.

Magnitude of $| \vec{v} |$: $\sqrt{{\left(3\right)}^{2} + {\left(- 2\right)}^{2} + {\left(7\right)}^{2}} = \sqrt{9 + 4 + 49} = \sqrt{62}$

Magnitude of $| \vec{u} |$: $\sqrt{{\left(5\right)}^{2} + {\left(- 3\right)}^{2} + {\left(2\right)}^{2}} = \sqrt{25 + 9 + 4} = \sqrt{38}$

So now we have:
$35 = \sqrt{62} \times \sqrt{38} \cos \theta$
$\cos \theta = \frac{35}{\sqrt{62} \times \sqrt{38}}$
$\theta = {\cos}^{-} 1 \left(\frac{35}{\sqrt{62} \times \sqrt{38}}\right)$
That, according to the calculator, equals to 43.86.

Alternatively, you can use the sine formula, which utilizes the cross product instead of the dot product, and it is slight more complicated.