If three points A, B and C have position vectors `hati + x hatj + 3 hatk, 3 hati + 4 hatj + 7 hatk and y hati -2hatj - 5 hatk ` respectively are collinear, then (x, y) =

Options

A. (2,-3)
B. (-2,3)
C. (2,3)
D. (-2,-3)

Answer

A

Explanation

Let `OA= hati+ x hatj + 3hatk`
`B= 3hati+4hatj+7 hatk and `
OC `= y hati -2hatj-5hatk`
Since AB,C are collinear.
then `AB= lambda BC`
`rArr 2hati+(4-hati)hatj+4hatk= lambda [ (y-3)hati-6-12hatk]`
On comparing the coefficient of `hati, hatj, and hatk` we, get
`2=(y-3)lambda " "...(i)`
`4-x=-6 lambda " "...(ii)`
and `4=-12 lambda`
`rArr lambda=-(1)/(3)`
On putting the value of `lambda` in Eq.a (i) and (i) we get,
`y=-3 and x=2`