JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Three Dimensional Geometry

Let

L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

and

L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}

be two lines. Then which of the following points lies on the line of the shortest distance between

L_{1}

and

L_{2}

Answer: (D)

Solution

Points on the two lines can be written as

P = (1 + 2 λ, 2 + 3 λ, 3 + 4 λ) on L_{1}, Q = (2 + 3 μ, 4 + 4 μ, 5 + 5 μ) on L_{2} .

The line of shortest distance meets both lines at right angles, so

P Q

must be perpendicular to both direction vectors

d_{1} = (2, 3, 4)

and

d_{2} = (3, 4, 5)

P Q = (3 μ - 2 λ + 1, 4 μ - 3 λ + 2, 5 μ - 4 λ + 2) .

P Q \cdot d_{1} = 0

2 (3 μ - 2 λ + 1) + 3 (4 μ - 3 λ + 2) + 4 (5 μ - 4 λ + 2) = 0 ⟹ 38 μ - 29 λ + 16 = 0. (1)

P Q \cdot d_{2} = 0

3 (3 μ - 2 λ + 1) + 4 (4 μ - 3 λ + 2) + 5 (5 μ - 4 λ + 2) = 0 ⟹ 50 μ - 38 λ + 21 = 0. (2)

Solving (1) and (2): from (1),

μ = \frac{29 λ - 16}{38}

; substituting into (2) and simplifying gives

6 λ = 2

, so

λ = \frac{1}{3}

, and then

μ = - \frac{1}{6}

P = (\frac{5}{3}, 3, \frac{13}{3}), Q = (\frac{3}{2}, \frac{10}{3}, \frac{25}{6}) .

Direction of

P Q

Q - P = (- \frac{1}{6}, \frac{1}{3}, - \frac{1}{6})

, i.e. proportional to

(1, - 2, 1)

.

Line of shortest distance:

\frac{x - \frac{5}{3}}{1} = \frac{y - 3}{- 2} = \frac{z - \frac{13}{3}}{1} .

Testing

(\frac{14}{3}, - 3, \frac{22}{3})

: each ratio gives

\frac{14/3 - 5/3}{1} = 3, \frac{- 3 - 3}{- 2} = 3, \frac{22/3 - 13/3}{1} = 3,

so this point lies on the line.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option