यदि `vecA = 3hati + 4hatj - 5hatk` तथा `vecB = 2hati + 3hatj - hatk` हो, तो (i) `vecA. vecB`, (ii) `vecA xx vecB` तथा (iii) `vecA` व `vecB` के बीच के कोण का मान ज्ञात कीजिए|

Explanation

(i) `vecA. vecB =(3hati + 4hatj - 5hatk).(2hati + 3hatj - hatk)`
`=3 xx 2 +4 + 5 xx 1`
`vecA. vecB= 23`
(ii) `vecA xx vecB = |{:(hati, hatj, hatk),(3,4,-5),(2,3,-1):}|`
`=hati(-4 + 15) - hatj (-3 + 10)+hatk(9-8)`
`=11hati - 7hatj + 3hatk`
(iii) `vecA` का परिमाण `A = sqrt(9+16+25) = 5sqrt(2)`
`vecB` का परिमाण `B=sqrt(4+9+1)=sqrt(14)`
`therefore vecA.vecB = AB cos theta` या `cos theta = (vecA.vecB)/(AB)` या `theta = cos^(-1)(vecA.vecB)/(vecA.vecB)`
`theta = cos^(-1)((23)/(5sqrt(28)))`