JEE Mains 2020 Model Paper 4 New Pattern

A(1,a,b);B(a,2,b);C(a,b,3)

→

AB

= (a−1)

^

i

+(2−a)

^

j

+0

^

k

→

BC

=

^

i

+(b−2)

^

j

+(3−b)

^

k

→

AB

= λ

→

BC

= λ(0

^

i

+(b−2)

^

j

+(3−b)

^

k

) where λ ≠ 0
Hence a - 1 = 0
⇒ a = 1 ... (i)
2 - a = λ (b - 2) ... (ii)
And 3 - b = 0 ⇒ b = 3 ... (iii)
With a = 1 and b = 3 , λ = 1
Hence a + b = 4 ⇒ (b)