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Three Dimensional Geometry: Mathematics | JEE Main

Consider the lines

L_1

and

L_2

given by

L_1

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

L_2

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

A line

L_3

having direction ratios

1, -1, -2

, intersects

L_1

and

L_2

at the points

P

and

Q

respectively. Then the length of line segment

PQ

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

3\sqrt{2}

4\sqrt{3}

4

2\sqrt{6}

Step-by-step solutionView

Correct answer

Express general points on lines $L_1$ and $L_2$ in terms of parameters $\lambda$ and $\mu$ , equate the direction ratios of the connecting segment to $\langle 1, -1, -2 \rangle$ , solve for the parameters to find points $P$ and $Q$ , and compute the distance $|PQ| = 2\sqrt{6}$ .

Option analysis

Why each option works or fails

A ·

3\sqrt{2}

The student miscalculates the distance formula using the components of $\vec{PQ}$ , computing $\sqrt{3^2 + 3^2} = 3\sqrt{2}$ by dropping the $z$ -coordinate difference. Ensure all three coordinate differences are squared and summed under the square root: $|PQ| = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{24} = 2\sqrt{6}$ .

B ·

4\sqrt{3}

The student introduces an arithmetic error when solving the system of equations for the parameters $\lambda$ and $\mu$ , arriving at an incorrect scaling factor of $2\sqrt{2}$ instead of $2$ , yielding $\sqrt{48} = 4\sqrt{3}$ . Carefully substitute the proportional relations $\frac{\mu - 2\lambda + 1}{1} = \frac{2\mu - \lambda - 1}{-1} = \frac{3\mu - 2\lambda + 1}{-2} = k$ to find $\lambda = 3$ and $\mu = 3$ , giving $k = -2$ and displacement vector $\langle 2, -2, -4 \rangle$ .

C ·

4

The student finds the scalar multiple relating the displacement vector $\vec{PQ}$ to its direction ratios or computes the value of a single coordinate difference (like $|\Delta z| = 4$ ) and mistakes it for the total length $|PQ|$ . Calculate the full Euclidean length of the segment $PQ$ rather than stopping at the absolute value of the largest coordinate difference.

D ·

2\sqrt{6}

None. The student correctly parameterized points $P$ and $Q$ , solved the system of equations for the parameters, and computed the distance $|PQ| = 2\sqrt{6}$ . None.

Reviewed route

Solution

StepWorking

01given

Line $L_1$ : $\frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2} = \lambda$ , Line $L_2$ : $\frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3} = \mu$ . Line $L_3$ passes through $P \in L_1$ and $Q \in L_2$ with direction ratios $(1, -1, -2)$ .

02goal

Find the coordinates of $P$ and $Q$ , then compute the Euclidean distance $|PQ|$ .

03approach

Express general points $P$ on $L_1$ and $Q$ on $L_2$ in terms of parameters $\lambda$ and $\mu$ . Find direction ratios of $\vec{PQ}$ and equate their ratios to the given direction ratios $(1, -1, -2)$ . Solve the resulting linear system for $\lambda$ and $\mu$ , determine $P$ and $Q$ , and compute distance $PQ$ .

04execute

Points are $P(2\lambda+1, \lambda+3, 2\lambda+2)$ and $Q(\mu+2, 2\mu+2, 3\mu+3)$ . The vector $\vec{QP} = P - Q = (2\lambda - \mu - 1, \lambda - 2\mu + 1, 2\lambda - 3\mu - 1)$ .

05execute

Equating proportional direction ratios: $\frac{2\lambda - \mu - 1}{1} = \frac{\lambda - 2\mu + 1}{-1} = \frac{2\lambda - 3\mu - 1}{-2} = k$ From the first pair: $-(2\lambda - \mu - 1) = \lambda - 2\mu + 1 \implies -2\lambda + \mu + 1 = \lambda - 2\mu + 1 \implies 3\lambda = 3\mu \implies \lambda = \mu$ . Substitute $\lambda = \mu$ into $\frac{2\lambda - \lambda - 1}{1} = \frac{2\lambda - 3\lambda - 1}{-2} \implies \lambda - 1 = \frac{-\lambda - 1}{-2} = \frac{\lambda + 1}{2}$ . Thus $2(\lambda - 1) = \lambda + 1 \implies 2\lambda - 2 = \lambda + 1 \implies \lambda = 3$ . Hence $\mu = 3$ .

06execute

Substituting $\lambda = 3, \mu = 3$ gives: $P = (2(3)+1, 3+3, 2(3)+2) = (7, 6, 8)$ $Q = (3+2, 2(3)+2, 3(3)+3) = (5, 8, 12)$ Then $PQ = \sqrt{(7-5)^2 + (6-8)^2 + (8-12)^2} = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$ .

✓verify

Vector $\vec{PQ} = Q - P = (5-7, 8-6, 12-8) = (-2, 2, 4) = -2(1, -1, -2)$ , which aligns with the given direction ratios $(1, -1, -2)$ .

Hints that build this answer step by step

Let $P$ lie on $L_1$ with parameter $\lambda$ and $Q$ lie on $L_2$ with parameter $\mu$ . How are the coordinates of $P$ and $Q$ written parametrically?
$P(2\lambda + 1, \lambda + 3, 2\lambda + 2)$ and $Q(\mu + 2, 2\mu + 2, 3\mu + 3)$
The vector $\vec{PQ} = Q - P$ is proportional to the direction ratios $\langle 1, -1, -2 \rangle$ . Which values of $\lambda$ and $\mu$ satisfy this condition?
$\lambda = 3$ and $\mu = 3$
Using $\lambda = 3$ and $\mu = 3$ , what is the length of segment $PQ$ ?
$2\sqrt{6}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why can we set the ratios equal to a common constant

k

Because direction ratios only define a direction up to a non-zero scalar multiplier. The actual displacement vector $\vec{QP}$ is $k(1, -1, -2)$ .

Answer

Express general points on lines $L_1$ and $L_2$ in terms of parameters $\lambda$ and $\mu$ , equate the direction ratios of the connecting segment to $\langle 1, -1, -2 \rangle$ , solve for the parameters to find points $P$ and $Q$ , and compute the distance $|PQ| = 2\sqrt{6}$ .

Why each option works or fails

A: $3\sqrt{2}$ - The student miscalculates the distance formula using the components of $\vec{PQ}$ , computing $\sqrt{3^2 + 3^2} = 3\sqrt{2}$ by dropping the $z$ -coordinate difference. Ensure all three coordinate differences are squared and summed under the square root: $|PQ| = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2} = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{24} = 2\sqrt{6}$ .
B: $4\sqrt{3}$ - The student introduces an arithmetic error when solving the system of equations for the parameters $\lambda$ and $\mu$ , arriving at an incorrect scaling factor of $2\sqrt{2}$ instead of $2$ , yielding $\sqrt{48} = 4\sqrt{3}$ . Carefully substitute the proportional relations $\frac{\mu - 2\lambda + 1}{1} = \frac{2\mu - \lambda - 1}{-1} = \frac{3\mu - 2\lambda + 1}{-2} = k$ to find $\lambda = 3$ and $\mu = 3$ , giving $k = -2$ and displacement vector $\langle 2, -2, -4 \rangle$ .
C: $4$ - The student finds the scalar multiple relating the displacement vector $\vec{PQ}$ to its direction ratios or computes the value of a single coordinate difference (like $|\Delta z| = 4$ ) and mistakes it for the total length $|PQ|$ . Calculate the full Euclidean length of the segment $PQ$ rather than stopping at the absolute value of the largest coordinate difference.
D · correct: $2\sqrt{6}$ - None. The student correctly parameterized points $P$ and $Q$ , solved the system of equations for the parameters, and computed the distance $|PQ| = 2\sqrt{6}$ . None.

Step-by-step solution

given: Line $L_1$ : $\frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2} = \lambda$ , Line $L_2$ : $\frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3} = \mu$ . Line $L_3$ passes through $P \in L_1$ and $Q \in L_2$ with direction ratios $(1, -1, -2)$ .
goal: Find the coordinates of $P$ and $Q$ , then compute the Euclidean distance $|PQ|$ .
approach: Express general points $P$ on $L_1$ and $Q$ on $L_2$ in terms of parameters $\lambda$ and $\mu$ . Find direction ratios of $\vec{PQ}$ and equate their ratios to the given direction ratios $(1, -1, -2)$ . Solve the resulting linear system for $\lambda$ and $\mu$ , determine $P$ and $Q$ , and compute distance $PQ$ .
execute: Points are $P(2\lambda+1, \lambda+3, 2\lambda+2)$ and $Q(\mu+2, 2\mu+2, 3\mu+3)$ . The vector $\vec{QP} = P - Q = (2\lambda - \mu - 1, \lambda - 2\mu + 1, 2\lambda - 3\mu - 1)$ .
execute: Equating proportional direction ratios: $\frac{2\lambda - \mu - 1}{1} = \frac{\lambda - 2\mu + 1}{-1} = \frac{2\lambda - 3\mu - 1}{-2} = k$ From the first pair: $-(2\lambda - \mu - 1) = \lambda - 2\mu + 1 \implies -2\lambda + \mu + 1 = \lambda - 2\mu + 1 \implies 3\lambda = 3\mu \implies \lambda = \mu$ . Substitute $\lambda = \mu$ into $\frac{2\lambda - \lambda - 1}{1} = \frac{2\lambda - 3\lambda - 1}{-2} \implies \lambda - 1 = \frac{-\lambda - 1}{-2} = \frac{\lambda + 1}{2}$ . Thus $2(\lambda - 1) = \lambda + 1 \implies 2\lambda - 2 = \lambda + 1 \implies \lambda = 3$ . Hence $\mu = 3$ .
execute: Substituting $\lambda = 3, \mu = 3$ gives: $P = (2(3)+1, 3+3, 2(3)+2) = (7, 6, 8)$ $Q = (3+2, 2(3)+2, 3(3)+3) = (5, 8, 12)$ Then $PQ = \sqrt{(7-5)^2 + (6-8)^2 + (8-12)^2} = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$ .
verify: Vector $\vec{PQ} = Q - P = (5-7, 8-6, 12-8) = (-2, 2, 4) = -2(1, -1, -2)$ , which aligns with the given direction ratios $(1, -1, -2)$ .

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

Express general points on lines $L_1$ and $L_2$ in terms of parameters $\lambda$ and $\mu$ , equate the direction ratios of the connecting segment to $\langle 1, -1, -2 \rangle$ , solve for the parameters to find points $P$ and $Q$ , and compute the distance $|PQ| = 2\sqrt{6}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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