3D Geometry – Distance along a specific direction of point to line – JEE Main 23 January 2025 Shift 2

Solution:

Let the given line be $L_1$: $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$.
Let the point be $P(1, 4, 0)$.
We need to find the distance from $P$ to $L_1$ measured along a direction parallel to the line $\frac{x-1}{1}=\frac{y-4}{2}=\frac{z}{3}$.
The direction ratios of the path are $(1, 2, 3)$.

Equation of the line passing through $P(1, 4, 0)$ with direction $(1, 2, 3)$ is:
$$\frac{x-1}{1} = \frac{y-4}{2} = \frac{z-0}{3} = \lambda$$
General point on this line is $Q(\lambda+1, 2\lambda+4, 3\lambda)$.

For this point $Q$ to lie on line $L_1$, it must satisfy the equation of $L_1$:
$$\frac{(\lambda+1)-2}{2} = \frac{(2\lambda+4)-6}{3} = \frac{3\lambda-3}{4}$$
$$\frac{\lambda-1}{2} = \frac{2\lambda-2}{3} = \frac{3\lambda-3}{4}$$

From the first two parts:
$$3(\lambda-1) = 2(2\lambda-2)$$
$$3\lambda – 3 = 4\lambda – 4$$
$$\lambda = 1$$

Let’s verify with the third part:
If $\lambda = 1$, then $\frac{1-1}{2} = 0$, $\frac{2-2}{3} = 0$, $\frac{3-3}{4} = 0$.
The condition is satisfied.

So, the point of intersection $Q$ is obtained by putting $\lambda = 1$ in the coordinates:
$$Q(1+1, 2(1)+4, 3(1)) = (2, 6, 3)$$

The required distance is the distance between $P(1, 4, 0)$ and $Q(2, 6, 3)$.
$$PQ = \sqrt{(2-1)^2 + (6-4)^2 + (3-0)^2}$$
$$PQ = \sqrt{1^2 + 2^2 + 3^2}$$
$$PQ = \sqrt{1 + 4 + 9} = \sqrt{14}$$

Ans. (2)