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

##### 1 Answer
Jan 7, 2016

$62.401$.

#### Explanation:

Definitions : Let $A = \left({a}_{1} , {a}_{2} , \ldots . , {a}_{n}\right) \mathmr{and} B = \left({b}_{1} , {b}_{2} , \ldots . , {b}_{n}\right)$ be any 2 vectors in a real or complex finite dimensional vector space X. Then we define:

1. The Euclidean inner product (dot product) of A and B as the real or complex number given by $A \cdot B = {a}_{1} {b}_{1} + {a}_{2} {b}_{2} + \ldots \ldots + {a}_{n} {b}_{n}$.
2. The norm of A as the real or complex number given by $| | A | | = \sqrt{{a}_{1}^{2} + {a}_{2}^{2} + \ldots \ldots + {a}_{n}^{2}}$.

Applying these 2 definitions to the given 3 dimensional vectors we get :

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

$= \left(6 \times - 4\right) + \left(9 \times - 1\right) + \left(- 1 \times 4\right)$

$= - 24 - 9 - 4$

$= - 37$.

$| | A | | = | | \left(6 , 9 , - 1\right) | | = \sqrt{{6}^{2} + {9}^{2} + {1}^{2}} = \sqrt{118}$.

Similarly $| | B | | = \sqrt{33}$.

$\therefore A \cdot B - | | A | | | | B | | = - 37 - \left(\sqrt{118}\right) \left(\sqrt{33}\right)$

$= - 37 - \sqrt{118 \times 33}$

$= 62.401$.