UPSC NDA Math Model Paper 3

Let the given points be A (−1, −1, 2), B (2, K, 5), C (3, 11, 6).
Then

→

AB

=(2+1)

→

i

+(k+1)

→

j

+(5−2)

→

k

=3

→

i

+(k+1)

→

j

+3

→

k

And

→

AC

=(3+1)

→

i

+(3+1)

→

j

+(2−2)

→

k

=4

→

i

+4

→

j

+4

→

k

Now, if A,B and C are collinear, then

→

AB

=λ

→

AC

⇒3

→

i

+(

→

k

+1)

→

j

+3

→

k

=λ(4

→

i

+4

→

j

+4

→

k

)
Comparing the coefficient of vector i, we get
⇒ 3 = 4 λ
∴ λ = 3/4
Now, comparing the coefficient of vector j, we get
⇒ (k + 1) = 4 λ
⇒ k + 1 = 4 × (3/4)
⇒ k + 1 = 3
∴ k = 2