If A = {2,-7 ,5}, B = {5,-7,4} and C = A - B, what is the angle between A and C?

Apr 5, 2016

$\alpha = 90 , {02}^{o}$

Explanation:

$\text{a) find C=A-B}$

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

$C = \left(2 - 5\right) , \left(- 7 + 7\right) , \left(5 - 4\right)$

$C = \left\{- 3 , 0 , 1\right\}$

$\text{b) find the magnitude of A and C:}$

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

$| | A | | = \sqrt{{2}^{2} + {\left(- 7\right)}^{2} + {5}^{2}} = \sqrt{4 + 49 + 25} = \sqrt{78}$

$| | C | | = \sqrt{{C}_{x}^{2} + {C}_{y}^{2} + {C}_{z}^{2}}$

$| | C | | = \sqrt{{\left(- 3\right)}^{2} + {0}^{2} + {1}^{2}} = \sqrt{9 + 0 + 1} = \sqrt{10}$

$\text{c) find dot product of A.C}$

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

$A \cdot C = 2 \cdot \left(- 3\right) + \left(- 7\right) \cdot 0 + 5 \cdot 1$

$A \cdot C = - 6 + 0 + 5 = - 1$

$\text{d) use the dot product formula}$

$A . C = | | A | | \cdot | | C | | \cdot \cos \alpha$

$A \cdot C = - 1$
$| | A | | = \sqrt{78}$
$| | C | | = \sqrt{10}$

$- 1 = \sqrt{78} \cdot \sqrt{10} \cdot \cos \alpha$

$\cos \alpha = - \frac{1}{\sqrt{78} \cdot \sqrt{10}}$

$\cos \alpha = - \frac{1}{\sqrt{780}}$

$\cos \alpha = - \frac{1}{27 , 93}$

$\alpha = 90 , {02}^{o}$