# If A = <7 ,-5 ,6 >, B = <4 ,-8 ,9 > and C=A-B, what is the angle between A and C?

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Oct 13, 2016

#### Answer:

theta ~~ 102.7°

#### Explanation:

Compute vector C:

$\overline{C} = \overline{A} - \overline{B} = \left(7 - 4\right) \hat{i} + \left(- 5 - - 8\right) \hat{j} + \left(6 - 9\right) \hat{k}$

$\overline{C} = 3 \hat{i} + 3 \hat{j} - 3 \hat{k}$

Compute the dot-product of vectors A and C:

$\overline{A} \cdot \overline{C} = \left(7\right) \left(3\right) + \left(- 5\right) \left(3\right) + \left(6\right) \left(- 3\right)$

$\overline{A} \cdot \overline{C} = \left(7\right) \left(3\right) + \left(- 5\right) \left(3\right) + \left(6\right) \left(- 3\right)$

$\overline{A} \cdot \overline{C} = - 12$

There is another form for the equation of the dot-product that contains the angle between the two vectors:

$\overline{A} \cdot \overline{C} = | \overline{A} | | \overline{C} | \cos \left(\theta\right) = - 12$

Compute the magnitude of vector A:

$| \overline{A} | = \sqrt{{7}^{2} + {\left(- 5\right)}^{2} + {6}^{2}}$

$| \overline{A} | = \sqrt{110}$

Compute the magnitude of vector C:

$| \overline{C} | = \sqrt{{3}^{2} + {3}^{2} + {\left(- 3\right)}^{2}}$

$| \overline{C} | = 3 \sqrt{3}$

$\sqrt{110} \left(3 \sqrt{3}\right) \cos \left(\theta\right) = - 12$

cos(theta) = -12/(sqrt(110)(3sqrt(3))

$\theta = {\cos}^{-} 1 \left(- \frac{4}{\sqrt{330}}\right)$

theta ~~ 102.7°

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