Show the equation of the points A (-7 , 4, - 2) , B (-2, 1,0) and
C (3, -2, 2) are collinear .

Explanation

If ` bara, barb and barc ` are position vectors of the points A , B
and C respectively , then
` bara = - 7hati + 4hatj - 2hatk `
`barb = - 2hati + hatj + 0hatk`
` barc = 3hatj - 2hatj + 2hatk `
` therefore bar(AB) = barb - bara `
` = 5hati - 3hatj + 2hatk `
` bar(AC) = barc - bara `
` = 10hati - 6hatj + 4hatk `
` = 2 [5hati - 3hatj + 2hatk]`
`bar(AC) = (2) bar(AB)` ...From (i)
That is , `bar(AC) ` is a scalar multiple of ` bar(AB)` . Hence, they
are collinear . But point A is in common . Hence , the point A , B and C are collinear .