# How do you use the definition of the scalar product, find the angles between the following pairs of vectors: 2i - 3 j k and - 2i - j - k?

Aug 1, 2016

${102.6}^{o}$

#### Explanation:

we use the definition of the scalar dot product, namely

$\vec{a} \cdot \vec{b} = \left\mid \vec{a} \right\mid \left\mid \vec{b} \right\mid \cos \varphi$

doing the LHS first [check the typo in the question, BTW]

$\left(\begin{matrix}2 \\ - 3 \\ 1\end{matrix}\right) \cdot \left(\begin{matrix}- 2 \\ - 1 \\ - 1\end{matrix}\right) = - 4 + 3 - 1 = - 2 q \quad \triangle$

next the RHS

$| \left(2\right) , \left(- 3\right) , \left(1\right) | | \left(- 2\right) , \left(- 1\right) , \left(- 1\right) | \cos \varphi$

$= \sqrt{{2}^{2} + {\left(- 3\right)}^{2} + {1}^{2}} \sqrt{{\left(- 2\right)}^{2} + {\left(- 1\right)}^{2} + {\left(- 1\right)}^{2}} \cos \varphi$

$= \sqrt{14} \sqrt{6} \cos \varphi q \quad \square$

comparing $\square$ and $\triangle$

$\sqrt{84} \cos \varphi = - 2$

$\cos \varphi = - \frac{2}{2 \sqrt{21}}$

$\varphi = {102.6}^{o}$