# If A= <-3 ,8 ,-1 > and B= <0 ,-4 ,2 >, what is A*B -||A|| ||B||?

$A \cdot B - | | A | | | | B | | = - 34 - 2 \cdot \sqrt{37 \cdot 10} \approx 72 , 5$

#### Explanation:

Our vectors are

$A = \left\langle - 3 , 8 , - 1\right\rangle ,$
$B = \left\langle 0 , - 4 , 2\right\rangle .$

Firstly, it's important to understand how the norm $| | \cdot | |$ is related to the inner product. By definition,

$| | A | {|}^{2} = A \cdot A .$

Therefore,

$A \cdot B - | | A | | | | B | | = A \cdot B - \sqrt{\left(A \cdot A\right) \left(B \cdot B\right)} .$

Calculating $A \cdot B$, $A \cdot A$ and $B \cdot B$, using the definition of the inner product in three dimensions, where ${A}_{i}$ is the $i$-th component of the vector $A = \left\langle {A}_{1} , {A}_{2} , {A}_{3}\right\rangle$,

$A \cdot B = {\sum}_{i = 1}^{3} {A}_{i} {B}_{i} ,$

$A \cdot B = - 3 \cdot 0 + 8 \cdot \left(- 4\right) + \left(- 1\right) \cdot 2 = - 34 ,$

$A \cdot A = {\left(- 3\right)}^{2} + {8}^{2} + {\left(- 1\right)}^{2} = 74 ,$

$B \cdot B = {0}^{2} + {\left(- 4\right)}^{2} + {2}^{2} = 20.$

Back to our expression,

A cdot B - sqrt((A cdot A) (B cdot B)) = -34 - sqrt(74 cdot 20)
A cdot B - sqrt((A cdot A) (B cdot B)) = -34 - 2 cdot sqrt(37 cdot 10)
A cdot B - sqrt((A cdot A) (B cdot B)) approx -72,5.

Therefore,

$A \cdot B - | | A | | | | B | | = - 34 - 2 \cdot \sqrt{37 \cdot 10} .$

Geometrically, this is a measure of how disaligned the two vectors are, since

$\frac{A \cdot B}{| | A | | | | B | |} - \frac{| | A | | | | B | |}{| | A | | | | B | |} = \cos \left(\theta\right) - 1 ,$

where $\theta$ is the angle between the vectors, and, therefore, the closer $A \cdot B - | | A | | | | B | |$ is to $0$, the more aligned are the vectors.