If $A = <2 ,4 ,-1 >$ , $B = <3 ,8 ,2 >$ and $C=A-B$ , what is the angle between A and C?

Question

1318 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-30T19:24:17

$theta ~~ 2.437$ radians

$C = <2-3, 4-8, -1-1>$

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

The rectangular definition of the dot-product is:

$A•C = (A_x)(C_x) + (A_y)(C_y) + (A_z)(C_z)$

$A•C = (2)(-1) + (4)(-4) + (-1)(-2)$

$A•C = -2 + -16 + 2$

$A•C = -16$

$|A| = sqrt(2^2 + 4^2 + (-1)^2)$

$|A| = sqrt(21)$

$|C| = sqrt((-1)^2 + (-4)^2 + (-2)^2)$

$|C| = sqrt(1 + 16 + 4)$

$|C| = sqrt(21)$

The polar definition of the dot-product is:

$A•C = |A||C|cos(theta)$

where $theta$ is the angle between the vectors.

Substituting what we have into the above:

$-16 = sqrt21sqrt21cos(theta)$

Solve for $theta$ :

$theta = cos^-1(-16/21)$

$theta ~~ 2.437$ radians

If A = <2 ,4 ,-1 >A = <2 ,4 ,-1 >, B = <3 ,8 ,2 >B = <3 ,8 ,2 > and C=A-BC=A-B, what is the angle between A and C?