यदि `vecA = hati + 2hatj -hatk` तथा `vecB =-hati + hatj -2hatk` हों, तो `vecA` व `vecB` के बीच कोण ज्ञात कीजिए |

Explanation

`vecA = hati + 2hatj -hatk` तथा `vecB = -hati + hatj -2hatk`
सदिशों `vecA` तथा `vecB` के अदिश परिमाण
`A = sqrt(A_(x)^(2)+A_(y)^(2)+A_(z)^(2))=sqrt((1)^(2)+(2)^(2)+(-1)^(2))=sqrt(6)`
`B=sqrt(B_(x)^(2)+B_(y)^(2)+B_(z)^(2))=sqrt((-1)^(2)+(1)^(2) + (-2)^(2)) = sqrt(6)`
`vecA * vecB = (hati + 2hatj -hatk)*(-hati + hatj - 2hatk) = -1 + 2 + 2 =3`
`vecA * vecB = AB cos theta`
`cos theta = (vecA * vecB)/(AB) = (3)/(sqrt(6)xxsqrt(6)) = (1)/(2)` अथवा `theta = 60^(@)`.