# What is the angle between <1,3,9 > and < 4,9,2 >?

Jan 13, 2016

The angle between the two vectors $= {59.2622}^{0}$

#### Explanation:

In order to find the angle between the two vectors, it helps to take the vectors in standard form (their tails are at the origin).Then take them as two sides of a triangle, and the vector of their difference would be the third side of that triangle.

So the triangle will have:

1. Side $\vec{A} = < 1 , 3 , 9 >$, the length of which is the magnitude of the vector$= | \vec{A} | = \sqrt{{\left(\sqrt{{1}^{2} + {3}^{2}}\right)}^{2} + {9}^{2}}$
$| \vec{A} | = \sqrt{91}$

2. Side $\vec{B} = < 4 , 9 , 2 >$, the length of which is the magnitude of the vector$= | \vec{B} | = \sqrt{{\left(\sqrt{{4}^{2} + {9}^{2}}\right)}^{2} + {2}^{2}}$
$| \vec{B} | = \sqrt{101}$

3. Side $\vec{C} = \vec{B} - \vec{A} = < 3 , 6 , - 7 >$, the length of which is the magnitude of the vector$= | \vec{C} | = \sqrt{{\left(\sqrt{{3}^{2} + {6}^{2}}\right)}^{2} + {\left(- 7\right)}^{2}}$
$| \vec{C} | = \sqrt{94}$

4. Three angles, one of which (the angle between $\vec{A}$and$\vec{B}$ $= \theta$) can be calculated using the law of cosines:
$\left(\cos \left(\theta\right) = \frac{| \vec{A} {|}^{2} + | \vec{B} {|}^{2} - | \vec{C} {|}^{2}}{2 | \vec{A} | \cdot | \vec{B} |}\right)$(MathIsFun, 2014)
$\cos \left(\theta\right) = \frac{91 + 101 - 94}{2 \cdot \sqrt{91} \cdot \sqrt{101}}$
$\cos \left(\theta\right) = 0.5111$
$\therefore \theta = {\cos}^{- 1} 0.5111$
$\theta = {59.2622}^{0}$

References:
-MathIsFun, 2014. The Law of Cosines . [Online]. Available from:https://www.mathsisfun.com/algebra/trig-cosine-law.html [Accessed:13th Jan 2016].