The position vectors of the points A,B and C are `(2hati + hatj - hatk), (3hati - 2hatj + hatk)` and `(hati + 4hatj - 3hatk)` respectively. Show that the points A,B and C are collinear.

Options

A. form an isosceles treangle
B. form a right angled triangle
C. are collinear
D. form a scalene triangle

Answer

C

Explanation

`vec(AB)=(3-2)hati+(-2-1)hatj+(1+1) hatk`
`= hati-3hati+2hatk`
`rArr |vec(AB)|= sqrt(1+9+4)=sqrt(14)`
`vec(BC) =(1-3)hati+(4+2)hatj+(-3-1)hatk`
`=-2hati+6hatj-4hatk`
` rArr |vec(AB)|=sqrt(4+36+16)`
`=sqrt(56)=2sqrt(14)`
`vec(CA)=(2-1)hati+(1-4)hatj+(-1+3)hatk`
`=hati-3hatj+2hatk`
`rArr |vec(CA)|= sqrt(1+9+4)=sqrt(14)`
So, `|vec(AB)|+|vec(AC)|=|vec(BC)|` and angle between AB and BC is `180^(@)`.
`:.` Points A,B,C cannot form an isosceles triangle,
Hence, A, B,C are collinear.