JEE Main 2024 (31 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Select Answer:

Progressive Hints

Hint 1

Visualize the geometry: You are dealing with skew lines in 3D space. They are like two airplanes flying at different altitudes—they never meet, and they aren't parallel. The shortest distance between them is the length of the unique line segment that is perpendicular to both of them simultaneously.

Hint 2

Recall your vector toolkit: To find the shortest distance, you need the 'common normal' vector. How do you find a vector perpendicular to two given vectors? Use the cross product. Once you have that normal vector, the shortest distance is simply the projection of the vector connecting any two points on the lines onto this normal vector.

Hint 3

The Execution: Start by defining the direction vectors

b_{1}

and

b_{2}

for both lines. Then, calculate the common normal

n = b_{1} \times b_{2}

. Finally, use the formula

d = \frac{∣ ( a _{1} - a _{2} ) \cdot n ∣}{∣ n ∣}

, where

a_{1}

and

a_{2}

are position vectors of points on

L_{1}

and

L_{2}

respectively.

Visualized Solution

Visualizing Skew Lines

We need to find the shortest distance between two skew lines, $L_{1}$ and $L_{2}$ .
Skew lines are non-parallel and non-intersecting lines in 3D space.
The shortest distance lies along their common perpendicular.

Extracting Data from $L_{1}$

Equation of $L_{1}$ : $\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$
Point on $L_{1}$ : $a_{1} = (1, - 1, - 4)$
Direction vector of $L_{1}$ : $b_{1} = (2, - 3, 2)$

Direction of Line $L_{2}$

$L_{2}$ passes through $A (- 4, 4, 3)$ and $B (- 1, 6, 3)$ .
Direction vector $b_{2} = A B = (- 1 - (- 4), 6 - 4, 3 - 3)$
$b_{2} = (3, 2, 0)$

Checking the Extra Condition

The problem states $L_{2}$ is perpendicular to $\frac{x - 3}{- 2} = \frac{y}{3} = \frac{z - 1}{1}$ .
Let's verify: $b_{2} \cdot b_{3} = (3) (- 2) + (2) (3) + (0) (1) = 0$ .
Since it's zero, the condition is satisfied. The two points $A$ and $B$ were sufficient!

Shortest Distance Formula

The shortest distance $d$ is the projection of the connecting vector onto the common perpendicular.
Formula: $d = \frac{∣ ( a _{1} - a _{2} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$

Calculating $a_{1} - a_{2}$

$a_{1} = (1, - 1, - 4)$ and $a_{2} = A = (- 4, 4, 3)$
$a_{1} - a_{2} = (1 - (- 4), - 1 - 4, - 4 - 3)$
$a_{1} - a_{2} = (5, - 5, - 7)$

Finding the Common Normal

We need a vector perpendicular to both $L_{1}$ and $L_{2}$ .
$b_{1} \times b_{2} = \hat{i} 23 \hat{j} - 3 2 \hat{k} 20$

Evaluating $b_{1} \times b_{2}$

$= \hat{i} (0 - 4) - \hat{j} (0 - 6) + \hat{k} (4 - (- 9))$
$= - 4 \hat{i} + 6 \hat{j} + 13 \hat{k}$
So, $b_{1} \times b_{2} = (- 4, 6, 13)$

Denominator: $∣ b_{1} \times b_{2} ∣$

$∣ b_{1} \times b_{2} ∣ = (- 4)^{2} + 6^{2} + 1 3^{2}$
$= 16 + 36 + 169$
$= 221$

Numerator: Dot Product

Numerator $= ∣ (a_{1} - a_{2}) \cdot (b_{1} \times b_{2}) ∣$
$= ∣ (5) (- 4) + (- 5) (6) + (- 7) (13) ∣$
$= ∣ - 20 - 30 - 91∣$
$= ∣ - 141∣ = 141$

Final Shortest Distance

Substitute the numerator and denominator back into the formula.
$d = \frac{141}{221}$
This is the shortest distance between the two skew lines.

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The Sigma Insight: Shortest Distance Between Two Skew Lines

Solution Diagram

Analyzing the Setup

To solve for the shortest distance between two skew lines, we first extract the geometric parameters of each line. For line

L_{1}

, given by the symmetric form:

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

We identify a point on the line

a_{1} = (1, - 1, - 4)

and its direction vector

b_{1} = (2, - 3, 2)

For line

L_{2}

, passing through points

A (- 4, 4, 3)

and

B (- 1, 6, 3)

, the direction vector

b_{2}

is calculated as

A B = (- 1 - (- 4), 6 - 4, 3 - 3) = (3, 2, 0)

. Note that any additional information regarding perpendicularity to other lines is a distractor and does not alter the fundamental direction of

L_{2}

The Common Normal

The shortest distance between two skew lines lies along a common normal vector, which is perpendicular to both

b_{1}

and

b_{2}

. We find this vector

n

using the cross product:

n = b_{1} \times b_{2} = \hat{i} 23 \hat{j} - 3 2 \hat{k} 20

Expanding the determinant, we obtain:

n = \hat{i} (0 - 4) - \hat{j} (0 - 6) + \hat{k} (4 - (- 9)) = - 4 \hat{i} + 6 \hat{j} + 13 \hat{k}

Thus, the common normal vector is

n = (- 4, 6, 13)

The Projection and Final Calculation

The shortest distance

d

is the projection of the vector connecting any two points on the lines onto the common normal vector. We define the vector connecting

a_{1}

and

a_{2}

(where

a_{2} = (- 4, 4, 3)

) as:

a_{1} - a_{2} = (1 - (- 4), - 1 - 4, - 4 - 3) = (5, - 5, - 7)

The distance formula is given by:

d = \frac{∣ ( a _{1} - a _{2} ) \cdot n ∣}{∣ n ∣}

Calculating the numerator:

∣ (5) (- 4) + (- 5) (6) + (- 7) (13) ∣ = ∣ - 20 - 30 - 91∣ = ∣ - 141∣ = 141

Calculating the denominator (the magnitude of

n

∣ n ∣ = (- 4)^{2} + 6^{2} + 1 3^{2} = 16 + 36 + 169 = 221

Combining these values, the shortest distance between the two skew lines is:

$d = \frac{141}{221}$

Common Traps & FAQs

Wait, why do we reject the negative root here?

Actually, we don't reject a root because we aren't solving a quadratic equation here! We are calculating a distance. Distance is a scalar quantity and must be positive. That is why we use the absolute value bars in our formula

∣ (a_{1} - a_{2}) \cdot (b_{1} \times b_{2}) ∣

. If your dot product comes out negative, the absolute value ensures your final distance is positive.

The problem mentions

L_{2}

is perpendicular to a third line. Do I need to use that?

That is a classic JEE distractor! In this specific problem, the two points

A

and

B

are sufficient to define the line

L_{2}

. The extra information about the third line is just a consistency check. If you calculate the dot product of your direction vector

b_{2}

with the direction vector of that third line, you will find it is zero, confirming the problem statement is consistent. You don't need to use it for the calculation.

Why can't we just use energy conservation or something simpler?

Because this is a problem of pure geometry, not dynamics! Energy conservation applies to moving particles. Here, we are analyzing the static structure of 3D space. The 'energy' of this problem lies in the elegance of the vector algebra. Stick to the vector projection method—it is the most robust and standard way to solve for the distance between skew lines.

Select Answer:

Visualized Solution

Visualizing Skew Lines

Extracting Data from $L_{1}$

Direction of Line $L_{2}$

Checking the Extra Condition

Shortest Distance Formula

Calculating $a_{1} - a_{2}$

Finding the Common Normal

Evaluating $b_{1} \times b_{2}$

Denominator: $∣ b_{1} \times b_{2} ∣$

Numerator: Dot Product

Final Shortest Distance

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Common Normal

The Projection and 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 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 $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ 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 $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is :

The line $l_{1}$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2 x + y - 2 z = 10$ . Then the shortest distance between the line $l_{1}$ and the line $\frac{x + 1}{2} = \frac{y + 4}{- 3} = \frac{z}{2}$ 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

The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}$ and $x + y + z + 1 = 0, 2 x - y + z + 3 = 0$ 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 $L_{2}$ ?

Select Answer:

Visualized Solution

Visualizing Skew Lines

Extracting Data from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

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

Checking the Extra Condition

Shortest Distance Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​​−a2​​

Finding the Common Normal

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​​×b2​​

Denominator: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣b1​​×b2​​∣

Numerator: Dot Product

Final Shortest Distance

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Common Normal

The Projection and Final Calculation

Similar Questions

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Extracting Data from $L_{1}$

Direction of Line $L_{2}$

Calculating $a_{1} - a_{2}$

Evaluating $b_{1} \times b_{2}$

Denominator: $∣ b_{1} \times b_{2} ∣$

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 $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

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

The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}$ and $x + y + z + 1 = 0, 2 x - y + z + 3 = 0$ is: