यदि `vecA=hati-hatj+2hatk` तथा `vecB=3hati+2hatj-hatk` तो ज्ञात कीजिए :
(i ) `vecA.vecB` तथा (ii ) `vecA` तथा `vecB` के बीच का कोण।

Explanation

प्रश्नानुसार, `" "vecA=hati-hatj+2hatk` तथा `vecB=3hati+2hatj-hatk`
(i) `vecA.vecB=(hati-hatj+2hatk).(3hati+2hatj-hatk)`
`=3(hati.hati)+2(hati.hatj)-(hati.hatk)-3(hatj.hati)-2(hatj.hatj)+(hatj.hatk)+6(hatk.hati)+4(hatk.hatj)-2(hatk.hatk)`
`=3-2-2=-1`
(ii)` |vecA|=sqrt((1)^(2)+(-1)^(2)+(2)^(2))=sqrt(1+1+4)=sqrt6`
`|vecB|=sqrt((3)^()+(2)^(2)+(-1)^(2))=sqrt(9+4+1)=sqrt(14)`
चूँकि `" "vecA.vecB=|vecA||vecB|cos theta`
`therefore" "cos theta=(vecA.vecB)/(|vecA||vecB|)=(-1)/(sqrt(6)sqrt(14))=(-1)/(sqrt(84))`
या `" "theta=cos^(-1)((-1)/(sqrt(84)))`