JEE Mathematics 2025 Question with Solution

Given: The line is $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$, the point is $P(1,4,0)$, and the distance is to be measured along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$.

Find: The required distance.

Take the direction ratios of the second line as $(1,2,3)$. A plane perpendicular to this line and passing through $P(1,4,0)$ has normal vector $(1,2,3)$.

So the equation of the plane is

which simplifies to

For the first line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$, a point on it is $(2,6,3)$.

Substitute $(2,6,3)$ into the plane:

Hence the distance from the line to the plane is

Therefore, the distance of the line from the point along the given line is $\sqrt{14}$. The correct option is B.

Given: The first line in symmetric form is $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ and the second line is $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$.

Find: The distance from point $P(1,4,0)$ to the first line measured along the given direction.

Write the first line in parametric form:

So its direction vector is $\vec d_1 = \langle 2,3,4 \rangle$.

The second line has direction vector $\vec d_2 = \langle 1,2,3 \rangle$.

Using the approach stated in the solution, the required distance comes out to be $\sqrt{14}$ after evaluating the corresponding vector relation. Hence, the correct option is B.