Question

A line with direction cosines proportional to 2, 1, 2 meets each of the lines $x = y + a = z$ and $x + a = 2 y = 2 z$. The co-ordinates of each of the points of intersection are given by

• A. (2a, 3a, 3a) (2a, a, a)
• B. (3a, 2a, 3a) (a, a, a)
• C. (3a, 2a, 3a) (a, a, 2a)
• D. (3a, 3a, 3a) (a, a, a)

Answer 1

Answer: (B)

(3a, 2a, 3a) (a, a, a)

Answer 2

Given lines are $\frac { x } { 1 } = \frac { y + a } { 1 } = \frac { z } { 1 } = \lambda$(say)

∴ Point is P(λ,λ – a, λ)

and $\frac { x + a } { 1 } = \frac { y } { 1 / 2 } = \frac { z } { 1 / 2 }$i.e. $\frac { x + a } { 2 } = \frac { y } { 1 } = \frac { z } { 1 } = \mu$ (say)

∴ Point Q(2μ – a, μ, μ)

Since d.r.’s of given lines are 2, 1, 2 and

d.r.’s of PQ = (2μ – a – λ, μ – λ + a, μ – λ)

According to question, $\frac { 2 \mu - a - \lambda } { 2 } = \frac { \mu - \lambda + a } { 1 } = \frac { \mu - \lambda } { 2 }$

Then $\lambda = 3 a$, μ = a.

Therefore, points of intersection are P(3a, 2a, 3a) and

Q(a, a, a).

Alternative method : Check by option $x = y + a = z$

i.e. $3 a = 2 a + a = 3 a$

⇒ a = a = a and $x + a = 2 y = 2 z$ i.e. $a + a = 2 a = 2 a$

⇒ a = a = a.

Hence (2) is correct