Maths Q14 – JEE-MAIN 2025

Line

L_{1}

passes through the point

(1, 2, 3)

and is parallel to

z

-axis. Line

L_{2}

passes through the point

(λ, 5, 6)

and is parallel to

y

-axis. Let for

λ = λ_{1}, λ_{2} < λ_{1}

, the shortest distance between the two lines be 3. Then the square of the distance of the point

(λ_{1}, λ_{2}, 7)

from the line

L_{1}

Clue 1 of 3: Representing Lines in 3D

Express the equations of lines $L_{1}$ and $L_{2}$ in vector form using the given point and direction vector.

🥷

Ninja Strategy•Geometric Simplification for Distance

Recognize that line $L_{1}$ is parallel to the z-axis. The distance of any point from $L_{1}$ depends only on its x and y coordinates relative to the line's x and y coordinates, making the calculation straightforward without needing the full cross-product formula for point-line distance.

🧠 Full Solution Path

Step 1: Determine

λ_{1}

and

λ_{2}

from Shortest Distance✦ Active

Line $L_{1}$ passes through $A_{1} = (1, 2, 3)$ and is parallel to the $z$ -axis, so its direction vector is $\vec{v_{1}} = (0, 0, 1)$ . Line $L_{2}$ passes through $A_{2} = (λ, 5, 6)$ and is parallel to the $y$ -axis, so its direction vector is $\vec{v_{2}} = (0, 1, 0)$ . The vector connecting points on the lines is $\vec{A_{1} A_{2}} = A_{2} - A_{1} = (λ - 1, 3, 3)$ . The cross product of direction vectors is $\vec{v_{1}} \times \vec{v_{2}} = (0, 0, 1) \times (0, 1, 0) = (- 1, 0, 0)$ . The magnitude is $| | \vec{v_{1}} \times \vec{v_{2}} | | = 1$ . The shortest distance $d$ is given by:

d = \frac{| \vec{A_{1} A_{2}} \cdot (\vec{v_{1}} \times \vec{v_{2}}) |}{| | \vec{v_{1}} \times \vec{v_{2}} | |} = \frac{| (λ - 1, 3, 3) \cdot (- 1, 0, 0) |}{1} = | - (λ - 1) | = | 1 - λ |

Given $d = 3$ , we have $| 1 - λ | = 3$ . This implies $1 - λ = 3$ or $1 - λ = - 3$ . Solving these gives $λ = - 2$ or $λ = 4$ . Since $λ_{2} < λ_{1}$ , we have $λ_{1} = 4$ and $λ_{2} = - 2$ .

Step 2: Calculate Distance of Point from Line

L_{1}

○ Expand

The point is $P = (λ_{1}, λ_{2}, 7) = (4, - 2, 7)$ . Line $L_{1}$ passes through $A_{1} = (1, 2, 3)$ and has direction vector $\vec{v_{1}} = (0, 0, 1)$ . The distance of point $P$ from line $L_{1}$ can be found by considering the projection of $\vec{A_{1} P}$ onto $\vec{v_{1}}$ . Alternatively, since $L_{1}$ is parallel to the $z$ -axis, its equation is $x = 1, y = 2, z = 3 + t$ . The distance of point $P (x_{0}, y_{0}, z_{0})$ from such a line is simply the distance in the $x y$ -plane.

D = \sqrt{(x_{0} - x_{1})^{2} + (y_{0} - y_{1})^{2}} = \sqrt{(4 - 1)^{2} + (- 2 - 2)^{2}}

Calculating the distance:

D = \sqrt{3^{2} + (- 4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5

💡 Teacher's Secret Hint

Remember that for a line parallel to an axis, the distance of a point from it simplifies significantly.

Step 3: Find the Square of the Distance○ Expand

The question asks for the square of the distance. The distance $D = 5$ .

D^{2} = 5^{2} = 25

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Maths Question 14 – JEE-MAIN 2025

Clue 1 of 3: Representing Lines in 3D

🧠 Full Solution Path