If A= <5 ,1 ,-2 > and B= <4 ,-2 ,3 >, what is A*B -||A|| ||B||?

May 12, 2017

$12 - \sqrt{870}$

Explanation:

For two vectors $A$ and $B$ of the form $A = < {A}_{x} , {A}_{y} , {A}_{z} >$ and $B = < {B}_{x} , {B}_{y} , {B}_{z} > ,$ the dot product $A \cdot B$ is given by $\left({a}_{x} \cdot {b}_{x}\right) + \left({a}_{y} \cdot {b}_{y}\right) + \left({a}_{z} \cdot {b}_{z}\right)$.

For $A = < 5 , 1 , - 2 >$ and $B = < 4 , - 2 , 3 >$, we have:

$A \cdot B = \left(5 \cdot 4\right) + \left(1 \cdot - 2\right) + \left(- 2 \cdot 3\right) = 20 - 2 - 6 = \textcolor{b l u e}{12}$

The next part is the product of the magnitudes of vectors $A$ and $B$. The magnitude of a vector $A = < {A}_{x} , {A}_{y} , {A}_{z} >$ is given by:

$| A | = \sqrt{{\left({A}_{x}\right)}^{2} + {\left({A}_{y}\right)}^{2} + {\left({A}_{z}\right)}^{2}}$

For the given vectors $A$ and $B$:

$| A | = \sqrt{{\left(5\right)}^{2} + {\left(1\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{25 + 1 + 4} = \sqrt{30}$

$| B | = \sqrt{{\left(4\right)}^{2} + {\left(- 2\right)}^{2} + {\left(3\right)}^{2}} = \sqrt{16 + 4 + 9} = \sqrt{29}$

We then have $| A | \cdot | B | = \sqrt{30} \cdot \sqrt{29} = \sqrt{870}$

Then $A \cdot B - | A | | B | = 12 - \sqrt{870}$.