If `veca=2veci+3vecj-4veck and vecb =5veci+2vecj+4veck`, find the angle between `veca and vecb`.

Answer

`cos^-1((38)/sqrt(1450))`

Explanation

Here `veca=2veci+3veci+4veck`
and`vecb=3veci+4vecj+5veck`
`veca.vecb=abcostheta`
`theta= cos^-1((veca.vecb)/(ab))`
=`cos^-1 (2xx3+3xx4+4xx5)/(sqrt(2^2+3^2+4^2)sqrt(3^2+4^2+^2))
=cos^-1 ((38)/(sqrt(29)sqrt(50)))

= cos^-1 ((38)/sqrt(1450))`