What is the angle between #<6,9,-2># and #< -2,0,4 >#?

Question

1289 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-12T17:20:05

#theta = arccos frac{ax + by + cz}{ sqrt {x^2 + y^2 + z^2} sqrt {a^2 + b^2 + c^2}}#

There are many multiplicities of this question, so let me solve all of them.

#v = (x, y, z)#

#w = (a, b, c)#

#v * w = |v| * |w| * cos theta#

#ax + by + cz = sqrt {x^2 + y^2 + z^2} sqrt {a^2 + b^2 + c^2} cos theta#

What is left is your arithmetical hability!

#-12 -8 = sqrt {36 + 81 + 4} sqrt {4 + 16} cos theta#

#-20 / (11 * sqrt20) = cos theta = - sqrt 20 / 11#