What is the angle between $<9,7,1>$ and $<8,3,6>$ ?

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Answer 1 · 2016-03-02T13:05:49

$alpha ~~34.06^@$

The angle $alpha$ between the two vectors

$vec u = < 9,7,1>$ and $vec v = <8,3,6>$

can be computed with the formula

$cos alpha = (vec u * vec v ) / (|| vec u || * || vec v ||)$

where the product in the numerator is a dot product.

In your case, this means:

$cos alpha = (9 * 8 + 7 * 3 + 1 * 6) / (sqrt(9^2 + 7^2 + 1^2) * sqrt(8^2 + 3^2 + 6^2)) = (99)/(sqrt(131)*sqrt(109)) ~~ 0.8285$

Thus,

$alpha ~~34.06^@$

What is the angle between <9,7,1><9,7,1> and <8,3,6> <8,3,6> ?