Find the angle `'theta'` between the vector `veca=2hati+3hatj-4hatk` and `vecb=3hati-2hatj+4hatk`.

Explanation

The angle `'theta'` between two vectors `veca` and `vecb` is given by
`cos theta=(veca.vecb)/(|veca||vecb|)` ltbrlt Now `" "veca.vecb=(2hai+3hatj-4hatk).(3hati-2hatj+4hatk)`
=6-6-16=-16
`|veca|=sqrt(2^(2)+3^(2)+4^(2))=sqrt(29)`
`|vecb|=sqrt(3^(2(-2)^(2)+4^(2))=sqrt(29)`
Therefore `cos theta=(-16)/(29)`
`rArr " " theta=cos^(-1)((-16)/(29))`