If $A = < 2, 3, - 4 >$ $A = <2 ,3 ,-4 >$ , $B = < 5, 1, 4 >$ $B = <5 ,1 ,4 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-10-20T03:45:43

$θ \approx 0.8$ $theta ~~ 0.8$ radians

To obtain the vector C, subtract the components of B from the respective components of a:

$C = < 2 - 5, 3 - 1, - 4 - 4 >$ $C = < 2 - 5, 3 - 1, -4 - 4>$

$C = < - 3, 2, - 8 >$ $C = < -3, 2, -8>$

There are two ways to compute $A \cdot C$ $A*C$ , the first is to add the products of the respective components of vector A and C:

$A \cdot C = (2) (- 3) + (3) (2) + (- 4) (- 8)$ $A*C = (2)(-3) + (3)(2) + (-4)(-8)$

$A \cdot C = 32$ $A*C = 32$

The second way is:

$A \cdot C = | A | | C | cos (θ)$ $A*C = |A||C|cos(theta)$

where $θ$ $theta$ is the angle between the two vectors.

We know the value of $A \cdot C$ $A*C$ but we need to compute |A| and |C|

$| A | = \sqrt{2^{2} + 3^{2} + {(- 4)}^{2}}$ $|A| = sqrt(2^2 + 3^2 + (-4)^2)$

$| A | = \sqrt{29}$ $|A| = sqrt(29)$

$| C | = \sqrt{{(- 3)}^{2} + 2^{2} + {(- 8)}^{2}}$ $|C| = sqrt((-3)^2 + 2^2 + (-8)^2)$

$| A | = \sqrt{77}$ $|A| = sqrt(77)$

$θ = {cos}^{- 1} (\frac{A \cdot C}{| A | | C |})$ $theta = cos^-1((A*C)/(|A||C|))$

$θ = {cos}^{- 1} (\frac{32}{\sqrt{29} \sqrt{77}})$ $theta = cos^-1((32)/(sqrt(29)sqrt(77)))$

$θ \approx 0.8$ $theta ~~ 0.8$ radians

If A = <2 ,3 ,-4 >A=<2,3,−4>A = <2 ,3 ,-4 >, B = <5 ,1 ,4 >B=<5,1,4>B = <5 ,1 ,4 > and C=A-BC=A−BC=A-B, what is the angle between A and C?