# If A= <2 ,1 ,-4 > and B= <3 ,2 ,7 >, what is A*B -||A|| ||B||?

##### 1 Answer
May 7, 2017

$A \cdot B - | \setminus | A | \setminus | | \setminus | B | \setminus |$ is asking for the dot product of A and B and the product of the magnitudes of vectors A and B.

For the dot product, we find the sum of the products for the x, y, and z terms, so we find that:
$A \cdot B = \left(2 \cdot 3\right) + \left(1 \cdot 2\right) + \left(- 4 \cdot 7\right) = 6 + 2 - 28 = - 20$

For the magnitudes, we use the distance formula for 3D vectors:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
or in this case, since the vectors already give the values of ${x}_{2} - {x}_{1} , {y}_{2} - {y}_{1} , {z}_{2} - {z}_{1}$, we simply plug in the given values:
$| \setminus | A | \setminus | = \sqrt{{2}^{2} + {1}^{2} + {\left(- 4\right)}^{2}} = \sqrt{21}$
$| \setminus | B | \setminus | = \sqrt{{3}^{2} + {2}^{2} + {7}^{2}} = \sqrt{62}$

By putting all the parts together, we get:
$A \cdot B - | \setminus | A | \setminus | | \setminus | B | \setminus | = - 20 - \sqrt{21} \cdot \sqrt{62} = - 20 - \sqrt{21 \cdot 62} = - 20 - \sqrt{1302} \approx - 56.083$