Let $P$ be the point of intersection of the lines $\dfrac{x-2}{1}=\dfrac{y-4}{5}=\dfrac{z-2}{1}$ and $\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{2}$. Then, the shortest distance of $P$ from the line $4x=2y=z$ is

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Three Dimensional Geometry

Let

P

be the point of intersection of the lines

\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}

and

\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}

. Then, the shortest distance of

P

from the line

4 x = 2 y = z

Answer: (C)

Line 1:

\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1} = r \Rightarrow (2 + r, 4 + 5 r, 2 + r)

. Line 2:

\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2} = s \Rightarrow (3 + 2 s, 2 + 3 s, 3 + 2 s)

.

Equating:

2 + r = 3 + 2 s, 4 + 5 r = 2 + 3 s, 2 + r = 3 + 2 s

(first and third are the same). From the first,

r = 1 + 2 s

; substitute into the second:

4 + 5 (1 + 2 s) = 2 + 3 s \Rightarrow 9 + 10 s = 2 + 3 s \Rightarrow 7 s = - 7 \Rightarrow s = - 1, r = - 1

P = (2 - 1, 4 - 5, 2 - 1) = (1, - 1, 1)

.

Line

4 x = 2 y = z = k \Rightarrow (k /4, k /2, k)

, direction

(1, 2, 4)

. Foot of perpendicular from P: minimise, or use

P Q \cdot (1, 2, 4) = 0

with

Q = (k /4, k /2, k)

(k /4 - 1) + 2 (k /2 + 1) + 4 (k - 1) = 0 \Rightarrow k /4 + k + 4 k - 1 + 2 - 4 = 0 \Rightarrow \frac{21 k}{4} = 3 \Rightarrow k = \frac{4}{7}

Q = (\frac{1}{7}, \frac{2}{7}, \frac{4}{7})

P Q = (1 - \frac{1}{7})^{2} + (- 1 - \frac{2}{7})^{2} + (1 - \frac{4}{7})^{2} = \frac{36 + 81 + 9}{49} = \frac{126}{49} = \frac{3 14}{7}

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option