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Three Dimensional Geometry: Mathematics | JEE Main

One vertex of a rectangular parallelopiped is at the origin

\mathrm{O}

and the lengths of its edges along

\mathrm{x}

\mathrm{y}

and

\mathrm{z}

axes are

3, 4

and

5

units respectively. Let

\mathrm{P}

be the vertex

(3, 4, 5)

. Then the shortest distance between the diagonal

\mathrm{OP}

and an edge parallel to

\mathrm{z}

axis, not passing through

\mathrm{O}

\mathrm{P}

is:

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.

\frac{12}{\sqrt{5}}

\frac{12}{5\sqrt{5}}

12\sqrt{5}

\frac{12}{5}

Step-by-step solutionView

Correct answer

The shortest distance between the diagonal $\mathrm{OP}$ and an edge parallel to the z-axis not passing through $\mathrm{O}$ or $\mathrm{P}$ is $\frac{12}{5}$ .

Option analysis

Why each option works or fails

A ·

\frac{12}{\sqrt{5}}

Believing that the denominator in the 2D distance formula involves $\sqrt{3^2 + 4^2} = 5$ under another radical, writing $\sqrt{5}$ instead of $5$ . Compute $\sqrt{a^2 + b^2} = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5$ , not $\sqrt{5}$ .

B ·

\frac{12}{5\sqrt{5}}

Dividing an extra factor of $\sqrt{5}$ coming from miscalculating the magnitude of the normal vector or cross product in 3D. In the standard 3D skew line formula, the denominator $|\vec{b}_1 \times \vec{b}_2|$ simplifies directly to $\sqrt{4^2 + 3^2} = 5$ , with no extra factor of $\sqrt{5}$ .

C ·

12\sqrt{5}

Multiplying the numerator by the hypotenuse instead of dividing by it when calculating the perpendicular distance from a point to a line. Remember that distance from $(x_0, y_0)$ to $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ , requiring division by $\sqrt{A^2 + B^2}$ .

D ·

\frac{12}{5}

None. This is the correct option. Projecting along the z-axis onto the xy-plane maps $\mathrm{OP}$ to the line $4x - 3y = 0$ and the target edges to $(3,0)$ or $(0,4)$ , yielding a perpendicular distance of $\frac{|4(3) - 3(0)|}{\sqrt{4^2 + (-3)^2}} = \frac{12}{5}$ .

Reviewed route

Solution

StepWorking

01given

A rectangular parallelopiped with vertex at $O(0,0,0)$ , edges along $x, y, z$ axes of lengths $3, 4, 5$ . Vertex $P = (3, 4, 5)$ . The target edge is parallel to the $z$ -axis, not passing through $O$ or $P$ .

02goal

Find the shortest distance between the line $OP$ and the selected edge parallel to the $z$ -axis.

03approach

Write the equations of the line $OP$ and the edge parallel to the $z$ -axis in vector form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ . Then apply the shortest distance formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ .

04execute

For line $OP$ : $\vec{a}_1 = (0,0,0)$ and $\vec{b}_1 = \langle 3, 4, 5 \rangle$ . For an edge parallel to the $z$ -axis not through $O$ or $P$ , choose the line passing through $(3, 0, 0)$ [or $(3,0,5)$ / $(0,4,0)$ ]: $\vec{a}_2 = (3, 0, 5)$ and $\vec{b}_2 = \langle 0, 0, 1 \rangle$ . Cross product of direction vectors: $\vec{b}_1 \times \vec{b}_2 = (3\hat{i} + 4\hat{j} + 5\hat{k}) \times \hat{k} = 4\hat{i} - 3\hat{j}$ , so $|\vec{b}_1 \times \vec{b}_2| = \sqrt{4^2 + (-3)^2} = 5$ . Dot product with $(\vec{a}_2 - \vec{a}_1) = 3\hat{i} + 0\hat{j} + 5\hat{k}$ : $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)| = |3(4) + 0(-3) + 5(0)| = 12$ . Shortest distance $d = \frac{12}{5}$ .

✓verify

The projection onto the $xy$ -plane reduces this to the distance from $(3,0)$ or $(0,4)$ to the line $4x - 3y = 0$ . The perpendicular distance from $(3,0)$ to $4x - 3y = 0$ is $\frac{|4(3) - 3(0)|}{\sqrt{4^2 + 3^2}} = \frac{12}{5}$ , which matches.

✓ 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

There are four edges parallel to the z-axis: passing through (0,0,0), (3,4,0), (3,0,0), and (0,4,0). Which one do we pick?

The edges through (0,0,0) and (3,4,0) pass through O and P respectively, which the problem explicitly excludes. The remaining two edges pass through (3,0,0) and (0,4,0). By symmetry of the rectangular box, both give the exact same shortest distance to OP.

Answer

The shortest distance between the diagonal $\mathrm{OP}$ and an edge parallel to the z-axis not passing through $\mathrm{O}$ or $\mathrm{P}$ is $\frac{12}{5}$ .

Why each option works or fails

A: $\frac{12}{\sqrt{5}}$ - Believing that the denominator in the 2D distance formula involves $\sqrt{3^2 + 4^2} = 5$ under another radical, writing $\sqrt{5}$ instead of $5$ . Compute $\sqrt{a^2 + b^2} = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5$ , not $\sqrt{5}$ .
B: $\frac{12}{5\sqrt{5}}$ - Dividing an extra factor of $\sqrt{5}$ coming from miscalculating the magnitude of the normal vector or cross product in 3D. In the standard 3D skew line formula, the denominator $|\vec{b}_1 \times \vec{b}_2|$ simplifies directly to $\sqrt{4^2 + 3^2} = 5$ , with no extra factor of $\sqrt{5}$ .
C: $12\sqrt{5}$ - Multiplying the numerator by the hypotenuse instead of dividing by it when calculating the perpendicular distance from a point to a line. Remember that distance from $(x_0, y_0)$ to $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ , requiring division by $\sqrt{A^2 + B^2}$ .
D · correct: $\frac{12}{5}$ - None. This is the correct option. Projecting along the z-axis onto the xy-plane maps $\mathrm{OP}$ to the line $4x - 3y = 0$ and the target edges to $(3,0)$ or $(0,4)$ , yielding a perpendicular distance of $\frac{|4(3) - 3(0)|}{\sqrt{4^2 + (-3)^2}} = \frac{12}{5}$ .

Step-by-step solution

given: A rectangular parallelopiped with vertex at $O(0,0,0)$ , edges along $x, y, z$ axes of lengths $3, 4, 5$ . Vertex $P = (3, 4, 5)$ . The target edge is parallel to the $z$ -axis, not passing through $O$ or $P$ .
goal: Find the shortest distance between the line $OP$ and the selected edge parallel to the $z$ -axis.
approach: Write the equations of the line $OP$ and the edge parallel to the $z$ -axis in vector form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ . Then apply the shortest distance formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ .
execute: For line $OP$ : $\vec{a}_1 = (0,0,0)$ and $\vec{b}_1 = \langle 3, 4, 5 \rangle$ . For an edge parallel to the $z$ -axis not through $O$ or $P$ , choose the line passing through $(3, 0, 0)$ [or $(3,0,5)$ / $(0,4,0)$ ]: $\vec{a}_2 = (3, 0, 5)$ and $\vec{b}_2 = \langle 0, 0, 1 \rangle$ . Cross product of direction vectors: $\vec{b}_1 \times \vec{b}_2 = (3\hat{i} + 4\hat{j} + 5\hat{k}) \times \hat{k} = 4\hat{i} - 3\hat{j}$ , so $|\vec{b}_1 \times \vec{b}_2| = \sqrt{4^2 + (-3)^2} = 5$ . Dot product with $(\vec{a}_2 - \vec{a}_1) = 3\hat{i} + 0\hat{j} + 5\hat{k}$ : $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)| = |3(4) + 0(-3) + 5(0)| = 12$ . Shortest distance $d = \frac{12}{5}$ .
verify: The projection onto the $xy$ -plane reduces this to the distance from $(3,0)$ or $(0,4)$ to the line $4x - 3y = 0$ . The perpendicular distance from $(3,0)$ to $4x - 3y = 0$ is $\frac{|4(3) - 3(0)|}{\sqrt{4^2 + 3^2}} = \frac{12}{5}$ , which matches.

Shortcut:

When to use it: Use this method when one line is parallel to a coordinate axis. Here, the line is parallel to the z-axis. Projecting the entire system along that axis onto the perpendicular plane collapses the line to a point.

given: Line $OP$ connects $(0,0,0)$ to $(3,4,5)$ . The edge is parallel to the $z$ -axis, passing through $(3,0,z)$ or $(0,4,z)$ .

approach: Project everything onto the $xy$ -plane along the $z$ -axis. The edge parallel to the $z$ -axis projects to the single point $(3,0)$ (or $(0,4)$ ). The line $OP$ projects to the line passing through $(0,0)$ and $(3,4)$ , which has equation $4x - 3y = 0$ . The 3D shortest distance equals the 2D perpendicular distance from the point to this line.

execute: Distance from point $(3,0)$ to the line $4x - 3y = 0$ is: $d = \frac{|4(3) - 3(0)|}{\sqrt{4^2 + (-3)^2}} = \frac{12}{5}$ .

verify: Testing the other valid edge $(0,4)$ : $d = \frac{|4(0) - 3(4)|}{\sqrt{4^2 + (-3)^2}} = \frac{12}{5}$ . Both give identical results.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The shortest distance between the diagonal $\mathrm{OP}$ and an edge parallel to the z-axis not passing through $\mathrm{O}$ or $\mathrm{P}$ is $\frac{12}{5}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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