# If A = <4 ,2 ,-5 >, B = <6 ,4 ,9 > and C=A-B, what is the angle between A and C?

Feb 22, 2016

$2.91 \text{ radians} , \left({166.8}^{\circ}\right)$

#### Explanation:

To calculate the angle between 2 vectors $\underline{a} \text{ and} \underline{c}$

use: $\cos \theta = \frac{\underline{a} . \underline{c}}{| \underline{a} | | \underline{c} |}$

where $\theta \text{ is the angle between" ulaand} \underline{c}$

here C = A - B = (4,2,-5) - (6,4,9) = (-2,-2,4)

hence: $\underline{a} . \underline{c} = \left(4 , 2 , - 5\right) . \left(- 2 , - 2 , 4\right)$

$= \left(4 \times - 2\right) + \left(2 \times - 2\right) + \left(- 5 \times 4\right) = - 8 - 4 - 20 = - 32$

$| \underline{a} | = \sqrt{{4}^{2} + {2}^{2} + {\left(- 5\right)}^{2}} = \sqrt{16 + 4 + 25} = \sqrt{45}$

and$| \underline{c} | = \sqrt{{\left(- 2\right)}^{2} + {\left(- 2\right)}^{2} + {4}^{2}} = \sqrt{4 + 4 + 16} = \sqrt{24}$

rArr theta = cos^-1((-32)/(sqrt45xxsqrt24)) ≈ 2.91" radians"