Question

find the angle between the following vectors:
A=12i-15j+25k , b=-12i+20j+30k
please show me steps and explanation :)

Answer 1

if you know the magnitude of each angle, then its just an application of the law of cosines

Answer 2

... for the law of cosines in this one, you would also need to know the length (magnitude) of the resultant vector from the sum of the other 2

Answer 3

snx :)

Answer 4

a= 12i-15j+25k
+ b=-12i+20j+30k
------------------
c = 0i + 5j +55k

Answer 5

once the lengths of each vector is known; the law of cosines can be constructed as:\[c^2=a^2+b^2-2ab~cos(\theta)\]
\[c^2-a^2-b^2=-2ab~cos(\theta)\]
\[\frac{c^2-a^2-b^2}{-2ab}=cos(\theta)\]
\[cos^{-1}\left(\frac{c^2-a^2-b^2}{-2ab}\right)=\theta\]

Answer 6

theres also a dot product version of this :)

Answer 7

\[cos~\theta=\frac{a.b}{|a|~|b|}\]
\[\theta=cos^{-1}\left(\frac{a.b}{|a|~|b|}\right)\]
and that is proved by the law of cosines

Answer 8

theta=1.3125