# What is the angle between <9,1,4>  and <6,-1,4 >?

Aug 2, 2018

$\theta = {\cos}^{-} 1 \left(\frac{69}{7 \sqrt{106}}\right)$

#### Explanation:

Let , $\vec{a} = < 9 , 1 , 4 > \mathmr{and} \vec{b} = < 6 , - 1 , 4 >$ be the two vectors.

$\therefore | \vec{a} | = \sqrt{{\left(9\right)}^{2} + {\left(1\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{81 + 1 + 16} = \sqrt{98} = 7 \sqrt{2}$

$\mathmr{and} | \vec{b} | = \sqrt{{\left(6\right)}^{2} + {\left(- 1\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{36 + 1 + 16} = \sqrt{53}$

Dot product : $a \cdot b = 9 \times 6 + 1 \times \left(- 1\right) + 4 \times 4$

$\therefore a \cdot b = 54 - 1 + 16 = 69$

Now the angle between $\vec{a} \mathmr{and} \vec{b}$ is:

$\theta = {\cos}^{-} 1 \left(\frac{a \cdot b}{| \vec{a} | | \vec{b} |}\right) = {\cos}^{-} 1 \left(\frac{69}{7 \sqrt{2} \sqrt{53}}\right)$

$\therefore \theta = {\cos}^{-} 1 \left(\frac{69}{7 \sqrt{106}}\right) =$${\left(16.78\right)}^{\circ}$