# If A = <3 ,2 ,5 >, B = <5 ,-7 ,8 > and C=A-B, what is the angle between A and C?

##### 1 Answer
Jun 25, 2016

$\theta = 87.12$

#### Explanation:

$\text{Strategy....}$

$1 \diamond \text{ find } C = A - B$
$2 \diamond \text{ find || A|| (magnitude of A)}$
$3 \diamond \text{ find ||C|| (magnitude of C)}$
$4 \diamond \text{ find A.C (dot product of A and C)}$
$5 \diamond \text{ use A.C=||A||.||C||. cos" theta " (dot product formula)}$

$1 \diamond \text{ ......................................................}$

${A}_{x} = 3 \text{ ; "A_y=2" ; } {A}_{z} = 5$

${B}_{x} = 5 \text{ ; "B_5=-7" ; } {B}_{z} = 8$

$C = < \left({A}_{x} - {B}_{x}\right) , \left({A}_{y} - {B}_{y}\right) , \left({A}_{z} - {B}_{z}\right) >$

$C = < \left(3 - 5\right) , \left(2 + 7\right) , \left(5 - 8\right)$

$\text{we obtain :}$
$C = < - 2 , 9 , - 3 >$

$2 \diamond \text{ ........................................................}$

$| | A | | = \sqrt{{A}_{x}^{2} + {A}_{y}^{2} + {A}_{z}^{2}} = \sqrt{{3}^{2} + {2}^{2} + {5}^{2}} = \sqrt{38}$

$3 \diamond \text{ ..........................................................}$

$| | C | | = \sqrt{{c}_{x}^{2} + {C}_{y}^{2} + {C}_{z}^{2}} = \sqrt{{\left(- 2\right)}^{2} + {9}^{2} + {\left(- 3\right)}^{2}}$

$| | C | | = \sqrt{4 + 81 + 9} = \sqrt{94}$

$4 \diamond \text{ ...........................................................}$

$A \cdot C = {A}_{x} \cdot {C}_{x} + {A}_{y} \cdot {C}_{y} + {A}_{z} \cdot {C}_{z}$

$A \cdot C = 3 \cdot \left(- 2\right) + 2 \cdot 9 + 5 \cdot \left(- 3\right) = - 6 + 18 - 15$

$A \cdot C = - 3$

$5 \diamond \text{ ............................................................}$

$- 3 = \sqrt{38} \cdot \sqrt{94} \cdot \cos \theta$

$\cos \theta = - \frac{3}{\sqrt{38 \cdot 94}}$

$\cos \theta = - \frac{3}{59.766}$

$\cos \theta = 0.0501957635$

$\theta = 87.12$