# If A= <2 ,-3 ,9 > and B= <-1, 3, 7 >, what is A*B -||A|| ||B||?

Aug 18, 2017

$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = 52 - \sqrt{5546}$
$\text{ } = - 22.4714710 \ldots$

#### Explanation:

We have:

$\boldsymbol{\underline{A}} = \left\langle2 , - 3 , 9\right\rangle$
$\boldsymbol{\underline{B}} = \left\langle- 1 , 3 , 7\right\rangle$

So then we can calculate the scalar product:

$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} = \left\langle2 , - 3 , 9\right\rangle \cdot \left\langle- 1 , 3 , 7\right\rangle$
$\text{ } = \left(2\right) \left(- 1\right) + \left(- 3\right) \left(3\right) + \left(9\right) \left(7\right)$
$\text{ } = - 2 - 9 + 63$
$\text{ } = 52$

And next the metric norms:

$| | \boldsymbol{\underline{A}} | | = \sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2} + {\left(9\right)}^{2}}$
$\text{ } = \sqrt{4 + 9 + 81}$
$\text{ } = \sqrt{94}$

$| | \boldsymbol{\underline{B}} | | = \sqrt{{\left(- 1\right)}^{2} + {\left(3\right)}^{2} + {\left(7\right)}^{2}}$
$\text{ } = \sqrt{1 + 9 + 49}$
$\text{ } = \sqrt{59}$

So the the result we require is:

$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = 52 - \sqrt{94} \sqrt{59}$
$\text{ } = 52 - \sqrt{5546}$
$\text{ } = - 22.4714710 \ldots$