यदि किसी त्रिभुज ABC के शीर्ष A , B , C क्रमशः (1 ,2 3) ,(-1 ,0 ,0 ) ,
(0 , 1 ,2 ) है तो `angle ABC ज्ञात कीजिए ।

Explanation

बिन्दुएँ A ,B तथा C के स्थिति सदिश क्रमशः `hati + 2hatj + 3hatk , - hati "तथा" hatj + 2hatk ` है
` therefore vec(OA) = hati + 2hatj + 3hatk `
`vec(OB) = - hati = -hati + 0hatj +0 hatk `
`vec(OC) = hatj + 2hatk = 0hati + hatj + 2hatk `
माना कि `angle ABC =theta "तो" cos theta = `(vec(BA) ।vec(BC))/(।vec(BA)।।vec(BC)।)` ...(1)
` [because angle ABC ,"सदिशों " vec(BA) " एवं" vec(BC)` के बीच का कोण है ]
अब `vec(BA)= vec(OA) - vec(OB) = (hati + 2hatj + 3hatk) - (-hati) = 2hati + 2hatj + 3hatk `
तथा अब `vec(BC) = vec(OC)- vec(OB) = (hatj + 2hatk) - (-hati)=hatj + 2hatk `
अब ` vec(BA).vec(BC) = 2xx1 + 2xx1 + 3xx2 = 10 `
` |vec(BA) | = sqrt(2^(2) + 2^(2) + 3^(2)) = sqrt(17)`
तथा ` |vec(BC) | = sqrt(1^(2) + 1^(2) + 2^(2)) = sqrt(6)`
(1) से, `cos theta = (10)/(sqrt(17 sqrt(6))) = (10)/(sqrt(102))= therefore = cos^(-1) (10)/(sqrt(102)) `
अतः `angle ABC =cos^(-1) ((10)/(sqrt(102)))`.