JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: The line $L_{1}$ is parallel to the vector $a = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$ and the line $L_{2}$ is parallel to the vector $b = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$ . The shortest distance between the lines $L_{1}$ and $L_{2}$ is :

Q: What if the lines were parallel? Would this formula still work?

Great catch! If the lines were parallel, $\vec{a}$ and $\vec{b}$ would be proportional, meaning their cross product would be the zero vector. The formula would result in division by zero, which is a mathematical signal that the formula is not applicable. In that case, you would use a different approach, specifically the distance from a point on one line to the other line.

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Progressive Hints

Hint 1

Hint 1: Imagine two lines in 3D space that never meet and are not parallel. These are 'skew lines'. To find the distance between them, you need to find a vector that is perpendicular to both lines simultaneously. What operation creates a vector perpendicular to two given vectors?

Hint 2

Hint 2: Once you have the common perpendicular vector, the shortest distance is simply the projection of the vector connecting any two points on the lines onto this common perpendicular direction. Recall the formula for the projection of a vector

v

onto a vector

n

Hint 3

Hint 3: Start by calculating the cross product

n = a \times b

to find the direction of the shortest distance. Then, compute the dot product of the displacement vector between the two given points with this normal vector, and divide by the magnitude of the normal vector.

Visualized Solution

Visualizing Skew Lines

Line $L_{1}$ passes through $P_{1} (7, 6, 2)$ and is parallel to $a = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ .
Line $L_{2}$ passes through $P_{2} (5, 3, 4)$ and is parallel to $b = 2 \hat{i} + \hat{j} + 3 \hat{k}$ .

The Shortest Distance Formula

Shortest Distance ( $S D$ ) between skew lines is given by:
$S D = \frac{∣ ( r _{2} - r _{1} ) \cdot ( a \times b ) ∣}{∣ a \times b ∣}$

Calculating $r_{2} - r_{1}$

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

Finding the Common Perpendicular Setup

$a \times b = \hat{i} - 3 2 \hat{j} 21 \hat{k} 43$

Evaluating $a \times b$

$= \hat{i} (6 - 4) - \hat{j} (- 9 - 8) + \hat{k} (- 3 - 4)$
$a \times b = 2 \hat{i} + 17 \hat{j} - 7 \hat{k}$

Calculating Magnitude $∣ a \times b ∣$

$∣ a \times b ∣ = 2^{2} + 1 7^{2} + (- 7)^{2}$
$∣ a \times b ∣ = 4 + 289 + 49 = 342$
$∣ a \times b ∣ = 338$

The Dot Product Calculation

$(r_{2} - r_{1}) \cdot (a \times b) = (- 2) (2) + (- 3) (17) + (2) (- 7)$
$= - 4 - 51 - 14 = - 69$

Final Substitution

$S D = \frac{∣ - 69∣}{3 38}$
$S D = \frac{69}{3 38} = \frac{23}{38}$

Conclusion

Final Answer: $\frac{23}{38}$

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The Sigma Insight: Shortest Distance Between Two Skew Lines

Solution Diagram

Analyzing the Setup

We are given two skew lines,

L_{1}

and

L_{2}

, in three-dimensional space.

L_{1}

passes through

P_{1} (7, 6, 2)

with direction vector

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

L_{2}

passes through

P_{2} (5, 3, 4)

with direction vector

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

. Our objective is to find the shortest distance between these two lines, which corresponds to the length of the common perpendicular segment.

The Power of the Cross Product

To find the direction of the shortest bridge, we calculate a vector

n

that is orthogonal to both

a

and

b

using the cross product:

n = a \times b = \hat{i} - 3 2 \hat{j} 21 \hat{k} 43

Expanding the determinant, we obtain:

n = \hat{i} (6 - 4) - \hat{j} (- 9 - 8) + \hat{k} (- 3 - 4) = 2 \hat{i} + 17 \hat{j} - 7 \hat{k}

This vector

n

defines the orientation of the shortest path between the two lines.

The Projection of Displacement

Next, we define the displacement vector between the two points

P_{1}

and

P_{2}

r_{2} - r_{1} = (5 - 7) \hat{i} + (3 - 6) \hat{j} + (4 - 2) \hat{k} = - 2 \hat{i} - 3 \hat{j} + 2 \hat{k}

The shortest distance (

S D

) is the scalar projection of this displacement vector onto the normal vector

n

. The formula is given by:

S D = \frac{∣ ( r _{2} - r _{1} ) \cdot n ∣}{∣ n ∣}

Final Calculation

First, we compute the dot product of the displacement vector and the normal vector:

(r_{2} - r_{1}) \cdot n = (- 2) (2) + (- 3) (17) + (2) (- 7) = - 4 - 51 - 14 = - 69

Next, we find the magnitude of the normal vector:

∣ n ∣ = 2^{2} + 1 7^{2} + (- 7)^{2} = 4 + 289 + 49 = 342 = 338

Substituting these values into our distance formula, we arrive at the final result:

S D = \frac{∣ - 69∣}{3 38} = \frac{69}{3 38} = \frac{23}{38}

The shortest distance between the two skew lines is $\frac{23}{38}$ units.

Common Traps & FAQs

Wait, why do we use the cross product here? I thought cross products were for areas of parallelograms.

That is a brilliant observation! While the magnitude of the cross product represents the area of a parallelogram, its direction is the key here. The cross product

a \times b

produces a vector that is orthogonal to both

a

and

b

. Since the shortest distance between two skew lines must be along a line perpendicular to both, this cross product gives us the exact direction we need to measure that 'gap' in 3D space.

Does it matter which point I pick on the lines? Can I pick any point?

Absolutely! The beauty of the shortest distance formula is that it relies on the vector connecting *any* two points on the respective lines. Because the lines are parallel to

a

and

b

, any displacement vector between the lines will have the same projection onto the common perpendicular. You are free to use the points provided, and the math will remain consistent.

Why do we take the absolute value in the numerator?

Distance, by definition, must be a non-negative scalar. The dot product can result in a negative value depending on the orientation of your vectors (the 'direction' of your measurement). By taking the absolute value, we ensure that we are measuring the magnitude of the separation, which is the physical distance we seek.

Is there a way to solve this without the formula? It feels like memorizing a black box.

You can absolutely derive it! You can define a general point on

L_{1}

P_{1} + t a

and on

L_{2}

P_{2} + s b

. The vector connecting these points is

(P_{2} - P_{1}) + s b - t a

. By setting the dot product of this vector with both

a

and

b

to zero, you can solve for

s

and

t

. It is a longer path, but it builds a much deeper intuition for how lines interact in space.

What if the lines were parallel? Would this formula still work?

Great catch! If the lines were parallel,

a

and

b

would be proportional, meaning their cross product would be the zero vector. The formula would result in division by zero, which is a mathematical signal that the formula is not applicable. In that case, you would use a different approach, specifically the distance from a point on one line to the other line.

Select Answer:

Visualized Solution

Visualizing Skew Lines

The Shortest Distance Formula

Calculating $r_{2} - r_{1}$

Finding the Common Perpendicular Setup

Evaluating $a \times b$

Calculating Magnitude $∣ a \times b ∣$

The Dot Product Calculation

Final Substitution

Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Power of the Cross Product

The Projection of Displacement

Final Calculation

Similar Questions

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The shortest distance between lines $L_{1}$ and $L_{2}$ , where $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$ and $L_{2}$ is the line passing through the points $A (- 4, 4, 3), B (- 1, 6, 3)$ and perpendicular to the line $\frac{x - 3}{- 2} = \frac{y}{3} = \frac{z - 1}{1}$ , is

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is :

Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is:

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ and be two lines. Then which of the following points lies on the line of the shortest distance between L₁ and $L_{2}$ ?

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is

Select Answer:

Visualized Solution

Visualizing Skew Lines

The Shortest Distance Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​−r1​

Finding the Common Perpendicular Setup

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a×b

Calculating Magnitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a×b∣

The Dot Product Calculation

Final Substitution

Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Power of the Cross Product

The Projection of Displacement

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 4x−3​=−11y+7​=5z−1​ and 3x−5​=−6y−9​=1z+2​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 2x−3​=−7y+15​=5z−9​ and 2x+1​=1y−1​=−3z−9​ is

Calculating $r_{2} - r_{1}$

Evaluating $a \times b$

Calculating Magnitude $∣ a \times b ∣$

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is :

Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is:

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is