# If A = <8 ,9 ,4 >, B = <6 ,7 ,4 > and C=A-B, what is the angle between A and C?

Aug 5, 2016

$= {18.67}^{o}$.
If the measurement is in the opposite sense, the angle is ${180}^{o} - {18.67}^{o} = {161.33}^{o}$

#### Explanation:

$C = < 8 , 9 , 4 > - < 6 , 7 , 4 > = < 2 , 2 , 0 >$.

Use the the formula

angle$= \theta = a r c \cos \frac{{a}^{2} + {c}^{2} - {b}^{2}}{2 a c} \mathmr{and} \pi - \theta$

$a = | < 8 , 9 , 4 \ge \sqrt{{8}^{2} + {9}^{2} + {4}^{2}} = \sqrt{161}$. Likewise,

$b = \sqrt{.} \left({6}^{2} + {7}^{2} + {4}^{2}\right) = \sqrt{101}$

$c = \sqrt{{2}^{2} + {2}^{2} + 0} = \sqrt{8}$

Now theta

$= a r c \cos \left(\frac{161 + 8 - 101}{2 X \sqrt{161} X \sqrt{8}}\right)$

$= {18.67}^{o}$

If the measurement is in the opposite sense, the angle is 180^o-

theta=161.33^o#

For checking the answer, you can use what I have used.

$\theta = a r c \sin \left(| A X C \frac{|}{| A | | C |} .\right)$