JEE Mathematics 2025 Question with Solution

Given:

$$L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$

and

$$L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$

Find: Which given point lies on the line of shortest distance between $L_1$ and $L_2$.

Let point $P$ on $L_1$ be

$$P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$$

and point $Q$ on $L_2$ be

$$Q(3\mu + 2, 4\mu + 4, 5\mu + 5)$$

Then the direction ratios of $PQ$ are

$$(3\mu - 2\lambda + 1, 4\mu - 3\lambda + 2, 5\mu - 4\lambda + 2)$$

Since the line of shortest distance is perpendicular to both lines, we use

$$PQ \cdot (2,3,4) = 0$$

and

$$PQ \cdot (3,4,5) = 0$$

So,

$$2(3\mu - 2\lambda + 1) + 3(4\mu - 3\lambda + 2) + 4(5\mu - 4\lambda + 2) = 0$$ $$38\mu - 29\lambda + 16 = 0$$

Also,

$$3(3\mu - 2\lambda + 1) + 4(4\mu - 3\lambda + 2) + 5(5\mu - 4\lambda + 2) = 0$$ $$50\mu - 38\lambda + 21 = 0$$

Solving these two equations,

$$\lambda = \frac{1}{3}, \qquad \mu = -\frac{1}{6}$$

Hence,

$$P = \left( \frac{5}{3}, 3, \frac{13}{3} \right)$$

and

$$Q = \left( \frac{3}{2}, \frac{10}{3}, \frac{25}{6} \right)$$

Therefore, direction ratios of $PQ$ are

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

So the equation of the line of shortest distance is

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

Now check option $D$, $\left( \frac{14}{3}, -3, \frac{22}{3} \right)$:

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

All three ratios are equal, so this point lies on the required line.

Therefore, the correct option is D.

Given: The line of shortest distance between two skew lines joins points on the two lines and is perpendicular to both.

Find: Which option lies on that line.

Take

$$P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \in L_1$$

and

$$Q(3\mu + 2, 4\mu + 4, 5\mu + 5) \in L_2$$

Then

$$\overrightarrow{PQ} = (3\mu - 2\lambda + 1, 4\mu - 3\lambda + 2, 5\mu - 4\lambda + 2)$$

Using perpendicularity with direction vectors $(2,3,4)$ and $(3,4,5)$, we get

$$\begin{aligned} 2(3\mu - 2\lambda + 1) + 3(4\mu - 3\lambda + 2) + 4(5\mu - 4\lambda + 2) &= 0 \\ 38\mu - 29\lambda + 16 &= 0 \end{aligned}$$

and

$$\begin{aligned} 3(3\mu - 2\lambda + 1) + 4(4\mu - 3\lambda + 2) + 5(5\mu - 4\lambda + 2) &= 0 \\ 50\mu - 38\lambda + 21 &= 0 \end{aligned}$$

From these,

$$\lambda = \frac{1}{3}, \qquad \mu = -\frac{1}{6}$$

Hence the required line is

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

Now testing the option points, the point

$$\left( \frac{14}{3}, -3, \frac{22}{3} \right)$$

satisfies all three equal ratios and hence lies on the line.

Therefore, the point on the line of shortest distance is $\left( \frac{14}{3}, -3, \frac{22}{3} \right)$, so the correct option is D.