# If A= <-5 ,-4 ,-7 > and B= <2 ,-9 ,8 >, what is A*B -||A|| ||B||?

Jun 21, 2018

$- 13260$

#### Explanation:

To answer this question we need three pieces:

• $A \cdot B$
• $| | A | |$
• $| | B | |$

but don't worry, they're all easy to compute!

• $A \cdot B$ is the scalar product of two vectors. As the name suggests, it is a number (scalar), computed as follows: $A$ and $B$ must have the same length, and we must add the products of the coordinates of $A$ and $B$ in the same positions.

So, if $A = \left({a}_{1} , {a}_{2} , \ldots , {a}_{n}\right)$ and $B = \left({b}_{1} , {b}_{2} , \ldots , {b}_{n}\right)$, we have

$A \cdot B = {a}_{1} {b}_{1} + {a}_{2} {b}_{2} + \ldots + {a}_{n} {b}_{n} = {\sum}_{i = 1}^{n} {a}_{i} {b}_{i}$

So, in your case, we have

$A \cdot B = - 5 \cdot 2 + \left(- 4\right) \left(- 9\right) + \left(- 7\right) \cdot 8 = - 10 + 36 - 56 = - 30$

The other pieces, $| | A | |$ and $| | B | |$, are called the norm of $A$ and $B$. Actually, the norm is defined as the scalar product of a vector with itself: $| | A | | = A \cdot A$, we already know how to compute them:

$| | A | | = \left(- 5\right) \left(- 5\right) + \left(- 4\right) \left(- 4\right) + \left(- 7\right) \left(- 7\right) = {\left(- 5\right)}^{2} + {\left(- 4\right)}^{2} + {\left(- 7\right)}^{2} = 25 + 16 + 49 = 90$

$| | B | | = {2}^{2} + {\left(- 9\right)}^{2} + {8}^{2} = 2 + 81 + 64 = 147$

$A \cdot B - | | A | | | B | | = - 30 - 90 \cdot 147 = - 30 - 13230 = - 13260$