What is the angle between $< - 4, 3, - 8 >$ $<-4,3,-8 >$ and $< 1, - 1, 5 >$ $< 1,-1,5>$ ?

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Answer 1 · 2016-10-08T14:06:32

$theta ~~ 2.56 radians or 146.76°$

The dot product is used, because it has two definitions:

$barA•barB = (A_1)(B_1) + (A_2)(B_2) + (A_3)(B_3)$

and

$barA•barB = |barA||barB|cos(theta)$

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

Compute the dot product, using the first definition:

$barA•barB = (-4)(1) + (3)(-1) + (-8)(5) = -41$

Compute the magnitude of vector A:

$|barA| = sqrt((-4)² + 3² + (-8)²) = sqrt(89)$

Compute the magnitude of vector B:

$|barB| = sqrt(1² + (-1)² + 5²) = sqrt(27)$

Substitute the above into the second equation:

$-41 = (sqrt(89))(sqrt(27))cos(theta)$

Solve for $theta$ :

$theta = cos^-1(-41/(sqrt(89)sqrt(27)))$

$theta ~~ 2.56 radians or 146.76°$

What is the angle between <-4,3,-8 ><−4,3,−8><-4,3,-8 > and < 1,-1,5><1,−1,5>< 1,-1,5>?