JEE MainMathematics

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Three Dimensional Geometry: Mathematics | JEE Main

Let the line

\mathrm{L}

pass through the point

(0,1,2)

, intersect the line

\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}

and be parallel to the plane

2 \mathrm{x}+\mathrm{y}-3 \mathrm{z}=4

. Then the distance of the point

\mathrm{P}(1,-9,2)

from the line

\mathrm{L}

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.

9

\sqrt{54}

\sqrt{69}

\sqrt{74}

Step-by-step solutionView

Correct answer

The distance from the point $(1, -9, 2)$ to the line $L$ is $\sqrt{74}$ .

Option analysis

Why each option works or fails

A ·

9

Calculated the distance along an unprojected displacement vector or mistakenly computed $\sqrt{81} = 9$ after an arithmetic error under the square root. Ensure the formula for the perpendicular distance $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$ or projection length is evaluated accurately without skipping vector components.

B ·

\sqrt{54}

Omitted the $z$ -coordinate difference when finding the perpendicular distance or made a sign error during the cross product. Double-check the evaluation of the cross product $\vec{AP} \times \vec{v} = (1, -10, 0) \times (-3, -1, -1) = (10, 1, -31)$ .

C ·

\sqrt{69}

Incorrectly solved the parallelism condition $\vec{d} \cdot \vec{n} = 0$ , obtaining an incorrect direction vector for $L$ . Remember that a line parallel to the plane $2x + y - 3z = 4$ must have its direction vector perpendicular to the normal vector $\langle 2, 1, -3 \rangle$ .

D ·

\sqrt{74}

None. This is the correct option. Correctly determined the point of intersection on the given line as $(-3, -4, -5)$ , giving direction vector $\langle 3, 1, 1 \rangle$ , and computed the distance as $\sqrt{74}$ .

Reviewed route

Solution

StepWorking

01given

Line $L$ passes through $A(0,1,2)$ , intersects the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at some point $B$ , and is parallel to the plane $2x+y-3z=4$ . We need the distance from $P(1,-9,2)$ to line $L$ .

02goal

Find the equation of line $L$ by determining the intersection point $B$ , then calculate the perpendicular distance from point $P$ to line $L$ .

03approach

Express the general point $B$ on the given line parametrically as $(1+2\lambda, 2+3\lambda, 3+4\lambda)$ . Then line $L$ along vector $\vec{AB}$ is parallel to the plane, so $\vec{AB} \cdot \vec{n}_{\text{plane}} = 0$ . Solve for $\lambda$ to get the direction of $L$ , parametrize point $Q$ on $L$ , and enforce $\vec{PQ} \perp L$ .

04execute

Let point $B = (1+2\lambda, 2+3\lambda, 3+4\lambda)$ . The direction vector of $L$ is $\vec{AB} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}$ . Since $L$ is parallel to the plane $2x+y-3z=4$ whose normal is $\vec{n} = 2\hat{i} + \hat{j} - 3\hat{k}$ , we have $\vec{AB} \cdot \vec{n} = 0$ : $2(1+2\lambda) + 1(1+3\lambda) - 3(1+4\lambda) = 0$ $2 + 4\lambda + 1 + 3\lambda - 3 - 12\lambda = 0 \implies -5\lambda = 0 \implies \lambda = 0.$ Thus, $\vec{AB} = \hat{i} + \hat{j} + \hat{k}$ . The equation of line $L$ is $\vec{r} = (0\hat{i} + 1\hat{j} + 2\hat{k}) + \mu(\hat{i} + \hat{j} + \hat{k})$ . A general point on $L$ is $Q(\mu, 1+\mu, 2+\mu)$ . For $Q$ to be the foot of perpendicular from $P(1, -9, 2)$ to $L$ , $\vec{PQ} \cdot (\hat{i}+\hat{j}+\hat{k}) = 0$ : $\vec{PQ} = (\mu - 1)\hat{i} + (10 + \mu)\hat{j} + \mu\hat{k}$ $(\mu - 1) + (10 + \mu) + \mu = 0 \implies 3\mu + 9 = 0 \implies \mu = -3.$ Therefore, $\vec{PQ} = -4\hat{i} + 7\hat{j} - 3\hat{k}$ . Distance $d = |\vec{PQ}| = \sqrt{(-4)^2 + 7^2 + (-3)^2} = \sqrt{16 + 49 + 9} = \sqrt{74}$ .

✓verify

Check that $Q(-3,-2,-1)$ satisfies line $L$ : $(-3-0)/1 = (-2-1)/1 = (-1-2)/1 = -3$ , which is consistent. Vector $\vec{PQ} = \langle -4, 7, -3 \rangle$ has dot product with direction $\langle 1, 1, 1 \rangle$ equal to $-4+7-3=0$ . Distance is $\sqrt{16+49+9} = \sqrt{74}$ .

Hints that build this answer step by step

Line $L$ passes through $A(0,1,2)$ and intersects the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at point $Q(2\lambda+1, 3\lambda+2, 4\lambda+3)$ . Since $L$ is parallel to the plane $2x+y-3z=4$ , which condition must $\lambda$ satisfy?
The vector $\vec{AQ}$ is perpendicular to the normal vector $\langle 2, 1, -3 \rangle$ , so $2(2\lambda+1) + 1(3\lambda+1) - 3(4\lambda+1) = 0$ .
Solving $2(2\lambda+1) + (3\lambda+1) - 3(4\lambda+1) = 0$ gives which value of $\lambda$ and direction vector for $L$ ?
$\lambda = 0$ , giving direction vector $\vec{d} = \langle 1, 1, 1 \rangle$
Now find the distance from $P(1, -9, 2)$ to the line $L$ passing through $A(0, 1, 2)$ with direction $\vec{d} = \langle 1, 1, 1 \rangle$ . What is this perpendicular distance?
$d = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|} = \frac{|\langle 1, -10, 0 \rangle \times \langle 1, 1, 1 \rangle|}{\sqrt{3}} = \frac{|\langle -10, -1, 11 \rangle|}{\sqrt{3}} = \frac{\sqrt{100+1+121}}{\sqrt{3}} = \sqrt{\frac{222}{3}} = \sqrt{74}$

✓ 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 is the direction vector perpendicular to the normal of the plane?

A line parallel to a plane lies entirely in a parallel plane. Therefore, its direction vector must be perpendicular to the plane's normal vector.

Answer

The distance from the point $(1, -9, 2)$ to the line $L$ is $\sqrt{74}$ .

Why each option works or fails

A: $9$ - Calculated the distance along an unprojected displacement vector or mistakenly computed $\sqrt{81} = 9$ after an arithmetic error under the square root. Ensure the formula for the perpendicular distance $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$ or projection length is evaluated accurately without skipping vector components.
B: $\sqrt{54}$ - Omitted the $z$ -coordinate difference when finding the perpendicular distance or made a sign error during the cross product. Double-check the evaluation of the cross product $\vec{AP} \times \vec{v} = (1, -10, 0) \times (-3, -1, -1) = (10, 1, -31)$ .
C: $\sqrt{69}$ - Incorrectly solved the parallelism condition $\vec{d} \cdot \vec{n} = 0$ , obtaining an incorrect direction vector for $L$ . Remember that a line parallel to the plane $2x + y - 3z = 4$ must have its direction vector perpendicular to the normal vector $\langle 2, 1, -3 \rangle$ .
D · correct: $\sqrt{74}$ - None. This is the correct option. Correctly determined the point of intersection on the given line as $(-3, -4, -5)$ , giving direction vector $\langle 3, 1, 1 \rangle$ , and computed the distance as $\sqrt{74}$ .

Step-by-step solution

given: Line $L$ passes through $A(0,1,2)$ , intersects the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at some point $B$ , and is parallel to the plane $2x+y-3z=4$ . We need the distance from $P(1,-9,2)$ to line $L$ .
goal: Find the equation of line $L$ by determining the intersection point $B$ , then calculate the perpendicular distance from point $P$ to line $L$ .
approach: Express the general point $B$ on the given line parametrically as $(1+2\lambda, 2+3\lambda, 3+4\lambda)$ . Then line $L$ along vector $\vec{AB}$ is parallel to the plane, so $\vec{AB} \cdot \vec{n}_{\text{plane}} = 0$ . Solve for $\lambda$ to get the direction of $L$ , parametrize point $Q$ on $L$ , and enforce $\vec{PQ} \perp L$ .
execute: Let point $B = (1+2\lambda, 2+3\lambda, 3+4\lambda)$ . The direction vector of $L$ is $\vec{AB} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}$ . Since $L$ is parallel to the plane $2x+y-3z=4$ whose normal is $\vec{n} = 2\hat{i} + \hat{j} - 3\hat{k}$ , we have $\vec{AB} \cdot \vec{n} = 0$ : $2(1+2\lambda) + 1(1+3\lambda) - 3(1+4\lambda) = 0$ $2 + 4\lambda + 1 + 3\lambda - 3 - 12\lambda = 0 \implies -5\lambda = 0 \implies \lambda = 0.$ Thus, $\vec{AB} = \hat{i} + \hat{j} + \hat{k}$ . The equation of line $2x+y-3z=4$ is $\vec{r} = (0\hat{i} + 1\hat{j} + 2\hat{k}) + \mu(\hat{i} + \hat{j} + \hat{k})$ . A general point on $L$ is $Q(\mu, 1+\mu, 2+\mu)$ . For $Q$ to be the foot of perpendicular from $P(1, -9, 2)$ to $L$ , $\vec{PQ} \cdot (\hat{i}+\hat{j}+\hat{k}) = 0$ : $\vec{PQ} = (\mu - 1)\hat{i} + (10 + \mu)\hat{j} + \mu\hat{k}$ $(\mu - 1) + (10 + \mu) + \mu = 0 \implies 3\mu + 9 = 0 \implies \mu = -3.$ Therefore, $\vec{PQ} = -4\hat{i} + 7\hat{j} - 3\hat{k}$ . Distance $d = |\vec{PQ}| = \sqrt{(-4)^2 + 7^2 + (-3)^2} = \sqrt{16 + 49 + 9} = \sqrt{74}$ .
verify: Check that $2 + 4\lambda + 1 + 3\lambda - 3 - 12\lambda = 0 \implies -5\lambda = 0 \implies \lambda = 0.$ satisfies line $L$ : $(-3-0)/1 = (-2-1)/1 = (-1-2)/1 = -3$ , which is consistent. Vector $\vec{PQ} = \langle -4, 7, -3 \rangle$ has dot product with direction $\langle 1, 1, 1 \rangle$ equal to $-4+7-3=0$ . Distance is $\sqrt{16+49+9} = \sqrt{74}$ .

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The distance from the point $(1, -9, 2)$ to the line $L$ is $\sqrt{74}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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