If A = <7 ,-5 ,4 >, B = <4 ,-8 ,-3 > and C=A-B, what is the angle between A and C?

Feb 11, 2016

First step is to find Vector $C$:
$C = A - B$
$= < 7 , - 5 , 4 >$-$< 4 , - 8 , - 3 > = < 3 , 3 , 7 >$

Angle between $A$ and $C$:

$\theta = {\cos}^{-} 1 \left(\frac{A \cdot C}{| A | | C |}\right) = {\cos}^{-} 1 \left(0.60\right) = {53}^{o}$

Explanation:

Vector $C$ is found as above, then the expression to find the angle, $\theta$, between $A$ and $C$ is:

$\theta = {\cos}^{-} 1 \left(\frac{A \cdot C}{| A | | C |}\right)$

The top line of this fraction is the vector dot product of the two vectors (which is itself a scalar, not a vector), and the bottom line is product of the lengths of the two vectors.

We are looking for the angle whose cosine is defined by these two numbers.

The dot product of two vectors:

$\left\{{a}_{1} , {a}_{2} , {a}_{3}\right\}$ $\cdot$ $\left\{{b}_{1} , {b}_{2} , {b}_{3}\right\} = \left({a}_{1} {b}_{1} + {a}_{2} {b}_{2} + {a}_{3} {b}_{3}\right)$

In this case:

$A \cdot B = < 7 , - 5 , 4 > \cdot < 4 , - 8 , - 3 > = \left(7 \cdot 4 + \left(- 5\right) \cdot 8 + \left(- 3\right) \cdot 7\right) = \left(28 + 40 - 21\right) = 47$

Finding the length of $A$ we use:

$l = \sqrt{{7}^{2} + {\left(- 5\right)}^{2} + {4}^{2}} = \sqrt{49 + 25 + 16} = \sqrt{90} = 9.5$

Using the same method, the length of $C$ is $\sqrt{67} = 8.2$

Putting it all together:

$\theta = {\cos}^{-} 1 \left(\frac{47}{9.5 \cdot 8.2}\right) = {53}^{o}$