# If A= <4, 9 ,-1 > and B= <9 ,-1 ,6 >, what is A*B -||A|| ||B||?

Aug 2, 2016

$A \cdot B - | | A | | | | B | | = - 86.54$

#### Explanation:

A= <4,9,-1)" ; "B=<9,-1,6>

$1 - \text{find } A \cdot B$

$A \cdot B = {A}_{x} \cdot {B}_{x} + {A}_{y} \cdot {B}_{y} + {A}_{z} \cdot {B}_{z}$

${A}_{x} = 4 \text{ ; } {B}_{x} = 9$
${A}_{y} = 9 \text{ ; } {B}_{y} = - 1$
${A}_{z} = - 1 \text{ ; } {B}_{z} = 6$

$A \cdot B = 4 \cdot 9 + 9 \cdot \left(- 1\right) + 6 \cdot \left(- 1\right)$

$A \cdot B = 36 - 9 - 6 = 36 - 15$
$A \cdot B = 21$

$2 - \text{find } | | A | |$
$| | A | | = \sqrt{{A}_{x}^{2} + {A}_{y}^{2} + {A}_{z}^{2}} = \sqrt{{4}^{2} + {9}^{2} + {\left(- 1\right)}^{2}}$
$| | A | | = \sqrt{16 + 81 + 1} = \sqrt{98}$

$3 - \text{find } | | B | |$

$| | B | | = \sqrt{{B}_{x}^{2} + {B}_{y}^{2} + {B}_{z}^{2}} = \sqrt{{9}^{2} + {\left(- 1\right)}^{2} + {6}^{2}}$
$| | B | | = \sqrt{81 + 1 + 36} = \sqrt{118}$

$4 - \text{find } | | A | | | | B | |$

$| | A | | | | B | | = \sqrt{98} \cdot \sqrt{118} = \sqrt{98 \cdot 118} = 107.54$

$5 - \text{find } A \cdot B - | | A | | | | B | |$

$A \cdot B - | | A | | | | B | | = 21 - 107.54$

$A \cdot B - | | A | | | | B | | = - 86.54$