यदि `veca=2hati+3hatj+4hatk` ओर `vecb=4hati+3hatj+2hatk,` तो `veca` तथा `vecb` के बिच का कोण निकाले।

Explanation

`veca*vecb=|veca||vecb|cos theta` या `cos theta""=(veca*vecb)/(|veca||vecb|).`
`veca*vecb=(2hati+3hatj+4hatk)*4hati+3hatj+2hatk)=8+9+8=25`
`|veca|^(2)=veca*veca=(2hati+3hatj+4hatk)*(2hati+3hatj+4hatk)=2^(2)+3^(2)+4^(2)=29.`
या `|veca|=sqrt29.`
`|vecb|^(2)=vecb*vecb=(4hati+3hatj+2hatk)*(4hati+3hatj+2hatk)=4^(2)+3^(2)+2^(2)=29.`
या `|vecb|=sqrt29.`
अतः `cos theta=(veca*vecb)/(|veca|*|vecb|)=(25)/(sqrt29*sqrt29)=25/29`
या `theta =cos ^(-1)((25)/(29)).`