If A = <4 ,2 ,5 >, B = <6 ,2 ,8 > and C=A-B, what is the angle between A and C?

Feb 19, 2016

2.82 radians≈ 162^@

Explanation:

To calculate the angle between 2 vectors $\underline{a} \text{ and } \underline{c}$
the following should be used.

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

here C= A - B = (4,2,5) - (6,2,8) = (-2,0,-3)

now $\underline{a} . \underline{c} = \left(4 , 2 , 5\right) . \left(- 2 , 0 , - 3\right)$

$= \left(4 \times - 2\right) + \left(2 \times 0\right) + \left(5 \times - 3\right)$

= -8 + 0 -15 = - 23

$| \underline{a} | = \sqrt{{4}^{2} + {2}^{2} + {5}^{2}} = \sqrt{16 + 4 + 25} = \sqrt{45}$

and $| \underline{c} | = \sqrt{{\left(- 2\right)}^{2} + 0 + {\left(- 3\right)}^{2}} = \sqrt{4 + 0 + 9} = \sqrt{13}$

rArr theta = cos^-1(-23/(sqrt45 xx sqrt13)) ≈ 2.82 " radians"