# If A= <7 ,-2 ,1 > and B= <-2 ,5 ,7 >, what is A*B -||A|| ||B||?

Oct 26, 2017

$\boldsymbol{\underline{A}} \boldsymbol{\cdot} \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = - 17 - 18 \sqrt{13}$

#### Explanation:

We have:

$\boldsymbol{\underline{A}} = \left\langle7 , - 2 , 1\right\rangle$
$\boldsymbol{\underline{B}} = \left\langle- 2 , 5 , 7\right\rangle$

So then the scalar product (or dot product) is:

$\boldsymbol{\underline{A}} \boldsymbol{\cdot} \boldsymbol{\underline{B}} = \left\langle7 , - 2 , 1\right\rangle \boldsymbol{\cdot} \left\langle- 2 , 5 , 7\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(7\right) \left(- 2\right) + \left(- 2\right) \left(5\right) + \left(1\right) \left(7\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 14 - 10 + 7$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 17$

And the moduli of the vectors are:

$| | \boldsymbol{\underline{A}} | | = \sqrt{{\left(7\right)}^{2} + {\left(- 2\right)}^{2} + {\left(1\right)}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{49 + 4 + 1}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{54}$

$| | \boldsymbol{\underline{B}} | | = \sqrt{{\left(- 2\right)}^{2} + {\left(5\right)}^{2} + {\left(7\right)}^{2}} =$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{4 + 25 + 49}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{78}$

And so:

$\boldsymbol{\underline{A}} \boldsymbol{\cdot} \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = - 17 - \sqrt{54} \sqrt{78}$
$\text{ } = - 17 - \sqrt{4212}$
$\text{ } = - 17 - 18 \sqrt{13}$