Question

The image of the point (1, 6, 3) in the line $\frac { x } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { 3 }$ is

• A. (1, 0, 7)
• B. (–1, 0, 7)
• C. (1, 0, –7)
• D. None of these

Answer 1

Answer: (A)

(1, 0, 7)

Answer 2

Let P(1, 6, 3) be the given point, and let L be the foot of the perpendicular from P to the given line. The co-ordinates of a general point on the given line are given by

$\frac { x - 0 } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { 3 } = \lambda$

i.e. $x = \lambda , y = 2 \lambda + 1 , z = 3 \lambda + 2$.

Let the co-ordinates of L be (λ, 2λ + 1, 3λ + 2) ……(i)

So, direction ratios of PL are $\lambda - 1,2 \lambda + 1 - 6,3 \lambda + 2 - 3$

i.e. $\lambda - 1,2 \lambda - 5,3 \lambda - 1$ .

Direction ratios of the given line are 1, 2, 3 which is perpendicular to PL.

∴ $( \lambda - 1 ) \cdot 1 + ( 2 \lambda - 5 ) 2 + ( 3 \lambda - 1 ) 3 = 0$ ⇒ $14 \lambda - 14 = 0$

⇒ $\lambda = 1$

So, co-ordinates of L are (1, 3, 5). Let $Q \left( x _ { 1 } , y _ { 1 } , z _ { 1 } \right)$ be the

image of $P ( 1,6,3 )$ in the given line.

Then L is the mid-point of PQ.

∴ $\frac { x _ { 1 } + 1 } { 2 } = 1 , \frac { y _ { 1 } + 6 } { 2 } = 3$ and $\frac { z _ { 1 } + 3 } { 2 } = 5$

⇒ $x _ { 1 } = 1 , y _ { 1 } = 0$ and $z _ { 1 } = 7$.

Hence the image of P(1, 6, 3) in the given line is (1, 0, 7).