JEE Main 2026 (21 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let the line $L_{1}$ be parallel to the vector $- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and pass through the point $(2, 6, 7)$ , and the line $L_{2}$ be parallel to the vector $2 \hat{i} + \hat{j} + 3 \hat{k}$ and pass through the point $(4, 3, 5)$ . If the line $L_{3}$ is parallel to the vector $- 3 \hat{i} + 5 \hat{j} + 16 \hat{k}$ and intersects the lines $L_{1}$ and $L_{2}$ at the points $C$ and $D$ , respectively, then $∣ C D ∣^{2}$ is equal to :

Q: Why do we use two different parameters, $\lambda_1$ and $\lambda_2$?

Excellent question. Because $C$ and $D$ are independent points on two different lines, their positions are governed by independent variables. If you used the same parameter, you would be forcing the points to move in lockstep, which is not physically true for these lines. By using $\lambda_1$ and $\lambda_2$, we allow each point to find its own unique location on its respective line.

Select Answer:

Visualized Solution

Equation of Line $L_{1}$

Line $L_{1}$ passes through $P_{1} (2, 6, 7)$ .
It is parallel to vector $v_{1} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ .

General Point $C$ on $L_{1}$

Let $C$ be any point on $L_{1}$ .
Using parameter $λ_{1}$ , $C = (2 - 3 λ_{1}, 6 + 2 λ_{1}, 7 + 4 λ_{1})$ .

Equation of Line $L_{2}$

Line $L_{2}$ passes through $P_{2} (4, 3, 5)$ .
It is parallel to vector $v_{2} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ .

General Point $D$ on $L_{2}$

Let $D$ be any point on $L_{2}$ .
Using parameter $λ_{2}$ , $D = (4 + 2 λ_{2}, 3 + λ_{2}, 5 + 3 λ_{2})$ .

Vector $C D$

The line $L_{3}$ intersects $L_{1}$ at $C$ and $L_{2}$ at $D$ .
Therefore, the vector connecting them is $C D = D - C$ .

Components of $C D$

$C D = (4 + 2 λ_{2} - (2 - 3 λ_{1})) \hat{i} + (3 + λ_{2} - (6 + 2 λ_{1})) \hat{j} + (5 + 3 λ_{2} - (7 + 4 λ_{1})) \hat{k}$
$C D = (2 + 2 λ_{2} + 3 λ_{1}) \hat{i} + (- 3 + λ_{2} - 2 λ_{1}) \hat{j} + (- 2 + 3 λ_{2} - 4 λ_{1}) \hat{k}$

Parallelism of $L_{3}$

Line $L_{3}$ is parallel to vector $v_{3} = - 3 \hat{i} + 5 \hat{j} + 16 \hat{k}$ .
Since $C D$ lies on $L_{3}$ , $C D ∥ v_{3}$ .

Direction Ratios Proportion

For parallel vectors, their direction ratios are proportional.
$\frac{2 + 2 λ _{2} + 3 λ _{1}}{- 3} = \frac{- 3 + λ _{2} - 2 λ _{1}}{5} = \frac{- 2 + 3 λ _{2} - 4 λ _{1}}{16}$

First Linear Equation

Equating the first two ratios:
$5 (2 + 2 λ_{2} + 3 λ_{1}) = - 3 (- 3 + λ_{2} - 2 λ_{1})$
$10 + 10 λ_{2} + 15 λ_{1} = 9 - 3 λ_{2} + 6 λ_{1}$
$9 λ_{1} + 13 λ_{2} = - 1 \dots (1)$

Second Linear Equation

Equating the last two ratios:
$16 (- 3 + λ_{2} - 2 λ_{1}) = 5 (- 2 + 3 λ_{2} - 4 λ_{1})$
$- 48 + 16 λ_{2} - 32 λ_{1} = - 10 + 15 λ_{2} - 20 λ_{1}$
$- 12 λ_{1} + λ_{2} = 38 \dots (2)$

Solving for $λ_{1}$ and $λ_{2}$

From (2): $λ_{2} = 12 λ_{1} + 38$
Substitute in (1): $9 λ_{1} + 13 (12 λ_{1} + 38) = - 1$
$9 λ_{1} + 156 λ_{1} + 494 = - 1 \Rightarrow 165 λ_{1} = - 495$
$λ_{1} = - 3$
$λ_{2} = 12 (- 3) + 38 = 2$

Exact Coordinates of $C$ and $D$

Substitute $λ_{1} = - 3$ into $C$ :
$C = (2 - 3 (- 3), 6 + 2 (- 3), 7 + 4 (- 3)) = (11, 0, - 5)$
Substitute $λ_{2} = 2$ into $D$ :
$D = (4 + 2 (2), 3 + 1 (2), 5 + 3 (2)) = (8, 5, 11)$

Exact Vector $C D$

$C D = (8 - 11) \hat{i} + (5 - 0) \hat{j} + (11 - (- 5)) \hat{k}$
$C D = - 3 \hat{i} + 5 \hat{j} + 16 \hat{k}$

Magnitude $∣ C D ∣^{2}$

We need to find the square of the magnitude of $C D$ .
$∣ C D ∣^{2} = (- 3)^{2} + 5^{2} + 1 6^{2}$
$∣ C D ∣^{2} = 9 + 25 + 256$
$∣ C D ∣^{2} = 290$
Final Answer: $290$

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

We are given two lines,

L_{1}

and

L_{2}

, in three-dimensional space. Our objective is to find the length of the segment

C D

that lies on a third line,

L_{3}

, which acts as a bridge between them.

Line

L_{1}

passes through

P_{1} (2, 6, 7)

with direction vector

v_{1} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}

. Any point

C

L_{1}

can be expressed using a parameter

λ_{1}

C = (2 - 3 λ_{1}, 6 + 2 λ_{1}, 7 + 4 λ_{1})

Line

L_{2}

passes through

P_{2} (4, 3, 5)

with direction vector

v_{2} = 2 \hat{i} + \hat{j} + 3 \hat{k}

. Any point

D

L_{2}

can be expressed using a parameter

λ_{2}

D = (4 + 2 λ_{2}, 3 + λ_{2}, 5 + 3 λ_{2})

The Master Equation

Line

L_{3}

is parallel to the vector

v_{3} = - 3 \hat{i} + 5 \hat{j} + 16 \hat{k}

. Since

L_{3}

contains the segment

C D

, the vector

C D

must be parallel to

v_{3}

We calculate the vector

C D

by subtracting the coordinates of

C

from

D

C D = (2 + 2 λ_{2} + 3 λ_{1}) \hat{i} + (- 3 + λ_{2} - 2 λ_{1}) \hat{j} + (- 2 + 3 λ_{2} - 4 λ_{1}) \hat{k}

Because

C D

is parallel to

v_{3}

, the ratios of their corresponding components must be equal:

\frac{2 + 2 λ _{2} + 3 λ _{1}}{- 3} = \frac{- 3 + λ _{2} - 2 λ _{1}}{5} = \frac{- 2 + 3 λ _{2} - 4 λ _{1}}{16}

Solving the System

This triple equality provides a system of two linear equations. Equating the first two ratios gives:

5 (2 + 2 λ_{2} + 3 λ_{1}) = - 3 (- 3 + λ_{2} - 2 λ_{1}) \Rightarrow 9 λ_{1} + 13 λ_{2} = - 1

Equating the last two ratios gives:

16 (- 3 + λ_{2} - 2 λ_{1}) = 5 (- 2 + 3 λ_{2} - 4 λ_{1}) \Rightarrow - 12 λ_{1} + λ_{2} = 38

Solving this system of linear equations yields the parameters:

λ_{1} = - 3, λ_{2} = 2

Final Calculation

Substituting these parameters back into our coordinate expressions, we find the points:

C = (2 - 3 (- 3), 6 + 2 (- 3), 7 + 4 (- 3)) = (11, 0, - 5)

D = (4 + 2 (2), 3 + 2, 5 + 3 (2)) = (8, 5, 11)

The vector

C D

is:

C D = D - C = (8 - 11) \hat{i} + (5 - 0) \hat{j} + (11 - (- 5)) \hat{k} = - 3 \hat{i} + 5 \hat{j} + 16 \hat{k}

The problem asks for the square of the length of the segment,

∣ C D ∣^{2}

∣ C D ∣^{2} = (- 3)^{2} + 5^{2} + 1 6^{2} = 9 + 25 + 256 = 290

The final result is 290.

Common Traps & FAQs

Why can't we just use the shortest distance formula between two skew lines?

That is a classic trap! The shortest distance formula is used when you are looking for the perpendicular distance between two lines that do not intersect. However, in this problem,

L_{3}

is a specific line that is given to intersect both

L_{1}

and

L_{2}

. We are not looking for the shortest distance; we are looking for the length of a specific segment defined by the intersection points

C

and

D

. Always read the problem carefully to see if the line is arbitrary or fixed!

Why do we use two different parameters,

λ_{1}

and

λ_{2}

Excellent question. Because

C

and

D

are independent points on two different lines, their positions are governed by independent variables. If you used the same parameter, you would be forcing the points to move in lockstep, which is not physically true for these lines. By using

λ_{1}

and

λ_{2}

, we allow each point to find its own unique location on its respective line.

Is there a faster way to solve the system of equations?

While you could use Cramer's rule or matrix inversion, for a

2 \times 2

system, substitution is often the safest and fastest way to avoid arithmetic errors. Stay calm, write clearly, and the algebra will resolve itself. Remember, in JEE Advanced, accuracy is just as important as speed.

Select Answer:

Visualized Solution

Equation of Line $L_{1}$

General Point $C$ on $L_{1}$

Equation of Line $L_{2}$

General Point $D$ on $L_{2}$

Vector $C D$

Components of $C D$

Parallelism of $L_{3}$

Direction Ratios Proportion

First Linear Equation

Second Linear Equation

Solving for $λ_{1}$ and $λ_{2}$

Exact Coordinates of $C$ and $D$

Exact Vector $C D$

Magnitude $∣ C D ∣^{2}$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

Solving the System

Final Calculation

Similar Questions

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The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

Select Answer:

Visualized Solution

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CD

Components of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CD

Parallelism of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

Direction Ratios Proportion

First Linear Equation

Second Linear Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ2​

Exact Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Exact Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CD

Magnitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣CD∣2

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Master Equation

Solving the System

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The square of the distance of the point (715​,732​,7) from the line 3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^​+7k^ is:

Equation of Line $L_{1}$

General Point $C$ on $L_{1}$

Equation of Line $L_{2}$

General Point $D$ on $L_{2}$

Vector $C D$

Components of $C D$

Parallelism of $L_{3}$

Solving for $λ_{1}$ and $λ_{2}$

Exact Coordinates of $C$ and $D$

Exact Vector $C D$

Magnitude $∣ C D ∣^{2}$

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is: