# If A = <6 ,4 ,-7 >, B = <3 ,5 ,0 > and C=A-B, what is the angle between A and C?

Mar 5, 2016

≈ 0.616" radians " (35.3^@)

#### Explanation:

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

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

where $\theta \text{ is the angle between the vectors }$

now C = A - B = (6,4,-7)-(3,5,0) = (3,-1,-7)

so $\underline{a} . \underline{c} = \left(6 , 4 , - 7\right) . \left(3 , - 1 , - 7\right)$

$= \left(6 \times 3\right) + \left(4 \times - 1\right) + \left(- 7 \times - 7\right) = 18 - 4 + 49 = 63$

$| \underline{a} | = \sqrt{{6}^{2} + {4}^{2} + {\left(- 7\right)}^{2}} = \sqrt{36 + 16 + 49} = \sqrt{101}$

$| \underline{c} | = \sqrt{{3}^{2} + {\left(- 1\right)}^{2} + {\left(- 7\right)}^{2}} = \sqrt{9 + 1 + 49} = \sqrt{59}$

rArr theta = cos^-1(63/(sqrt101xxsqrt59)) ≈ 0.616" radians "