CGPET 2011 Solved Paper

Given that the points with position vectors 60i+3j,40i−8j and i−52j are collinear, then there exist three scalars 1,x and y such that
‌1+x+y=0 . . . (i)
and(60i+3j)⋅1+(40i−8j)⋅x
+(ai−52j)⋅y=0
‌⇒‌‌(60+40x+ay)i+(3−8x−52y)j=0
‌‌=0i+0j

Comparing both sides, we get
‌‌60+40x+ay=0 . . . (ii)
and 3−8x−52y=0 . . . (iii)
Solving Eqs. (i) and (iii), we get
x=‌

−5

4

,y=‌

1

4

Then, from Eq. (ii),
60+40(‌

−5

4

)+a(‌

1

4

)‌=0
⇒‌‌‌

a

4

=−10⇒a‌=−40