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

Question

1488 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-20T03:45:43

$theta ~~ 0.8$ radians

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

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

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

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

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

$A*C = 32$

The second way is:

$A*C = |A||C|cos(theta)$

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

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

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

$|A| = sqrt(29)$

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

$|A| = sqrt(77)$

$theta = cos^-1((A*C)/(|A||C|))$

$theta = cos^-1((32)/(sqrt(29)sqrt(77)))$

$theta ~~ 0.8$ radians

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