JEE Main 2024 (27 Jan Shift 1)

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Animated Solution for Mathematics - Three Dimensional Geometry: The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :

Q: Is it possible for $\lambda$ to be a fraction?

Absolutely! In many JEE problems, the parameter $\lambda$ can be a fraction, a negative number, or even zero. Do not be alarmed if you get a non-integer value. In this specific problem, we found $\lambda = -3$, which is a clean integer, but always trust your algebraic steps regardless of the result.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Orient yourself. The problem asks for the distance of a point from a line, but it is NOT the shortest (perpendicular) distance. It is a distance measured along a specific direction. Visualize a vector starting at the point and ending on the line, which must be parallel to the second given line.

Hint 2

Hint 2 (The Arsenal): Use the parametric form of the target line

L_{1}

to define an arbitrary point

B

on it. If you set the line equation equal to a parameter

λ

, you can express the coordinates of

B

as a function of

λ

. Then, construct the vector

P B

by subtracting the coordinates of

P

from

B

Hint 3

Hint 3 (The Execution): Since the segment

P B

is parallel to the line

L_{2}

, their direction ratios must be proportional. Set the direction ratios of

P B

proportional to the direction ratios of

L_{2}

(which are

2, - 3, 6

). This will give you an equation to solve for

λ

. Once you have

λ

, find the coordinates of

B

and use the distance formula.

Visualized Solution

Visualizing the Problem

Point $P (7, - 2, 11)$
Target Line $L_{1}$ : $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$
Direction Line $L_{2}$ : $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$
We need to find the distance $P B$ where $B$ lies on $L_{1}$ and $P B ∥ L_{2}$ .

Defining a General Point $B$ on $L_{1}$

Let $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3} = λ$
General point $B$ on $L_{1}$ :
$x = λ + 6$ , $y = 4$ , $z = 3 λ + 8$
So, $B = (λ + 6, 4, 3 λ + 8)$

Understanding 'Distance Along a Line'

Distance measured along $L_{2}$ means segment $P B ∥ L_{2}$ .
Direction ratios (DRs) of $L_{2}$ are $(2, - 3, 6)$ .
Therefore, DRs of $P B$ must be proportional to $(2, - 3, 6)$ .

Finding Direction Ratios of $P B$

DRs of $P B = (x_{B} - x_{P}, y_{B} - y_{P}, z_{B} - z_{P})$
$x$ -component: $(λ + 6) - 7 = λ - 1$
$y$ -component: $4 - (- 2) = 6$
$z$ -component: $(3 λ + 8) - 11 = 3 λ - 3$
DRs of $P B$ : $(λ - 1, 6, 3 λ - 3)$

Setting up the Proportionality

Since $P B ∥ L_{2}$ , their DRs are proportional:
$\frac{λ - 1}{2} = \frac{6}{- 3} = \frac{3 λ - 3}{6}$
Simplify the middle ratio: $\frac{6}{- 3} = - 2$

Solving for $λ$

Equating the first term to the constant:
$\frac{λ - 1}{2} = - 2$
$λ - 1 = - 4$
$λ = - 3$

Finding the Coordinates of $B$

Substitute $λ = - 3$ into $B = (λ + 6, 4, 3 λ + 8)$ :
$x = - 3 + 6 = 3$
$y = 4$
$z = 3 (- 3) + 8 = - 1$
Point $B = (3, 4, - 1)$

Applying the Distance Formula

Distance $P B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Substitute $P (7, - 2, 11)$ and $B (3, 4, - 1)$ :
$P B = (7 - 3)^{2} + (- 2 - 4)^{2} + (11 - (- 1))^{2}$

Final Calculation

$P B = 4^{2} + (- 6)^{2} + 1 2^{2}$
$P B = 16 + 36 + 144$
$P B = 196$
$P B = 14$

Conclusion

Final Answer: $14$
Key Takeaway: 'Distance along a line' implies the path is parallel to the given direction vector.

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

We are given a point

P (7, - 2, 11)

and a target line

L_{1}

defined by the equation:

\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}

We seek the distance from

P

L_{1}

measured along a path parallel to the line

L_{2}

, which is defined by:

\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}

Phase 1

The Parametric Bridge

To find the distance, we must identify the intersection point

B

where the path from

P

meets

L_{1}

. Since

B

lies on

L_{1}

, we represent its coordinates using a parameter

λ

\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3} = λ

This yields the parametric coordinates for

B

B = (λ + 6, 4, 3 λ + 8)

Phase 2

The Vector Alignment

Next, we define the vector

P B

connecting the starting point

P (7, - 2, 11)

to the point

B

on the line

L_{1}

P B = ((λ + 6) - 7, 4 - (- 2), (3 λ + 8) - 11)

Simplifying the components, we obtain:

P B = (λ - 1, 6, 3 λ - 3)

Since the path is constrained to be parallel to

L_{2}

, the vector

P B

must be proportional to the direction ratios of

L_{2}

, which are

(2, - 3, 6)

. This leads to the following proportionality:

\frac{λ - 1}{2} = \frac{6}{- 3} = \frac{3 λ - 3}{6}

Phase 3

The Final Calculation

We observe that the middle term simplifies to a constant:

\frac{6}{- 3} = - 2

Equating the first term to this constant allows us to solve for

λ

\frac{λ - 1}{2} = - 2 \Rightarrow λ - 1 = - 4 \Rightarrow λ = - 3

Substituting

λ = - 3

back into our parametric expressions for

B

x = - 3 + 6 = 3, y = 4, z = 3 (- 3) + 8 = - 1

Thus, the intersection point is

B (3, 4, - 1)

. We now calculate the distance

P B

using the 3D distance formula:

P B = (7 - 3)^{2} + (- 2 - 4)^{2} + (11 - (- 1))^{2}

P B = 4^{2} + (- 6)^{2} + 1 2^{2} = 16 + 36 + 144 = 196

The final distance is $14$ .

Common Traps & FAQs

Wait, why can't we just use the perpendicular distance formula for a point and a line?

That is a fantastic question! The standard perpendicular distance formula is specifically designed for the shortest distance between a point and a line. However, this problem introduces a constraint: the distance must be measured *along* a specific direction (parallel to

L_{2}

). This means we are not looking for the shortest path, but a path that follows a specific vector. Always read the problem carefully—if it specifies a direction, you must respect that constraint!

How do I know which direction vector to use for the proportionality condition?

The problem states the distance is measured *along* the line

L_{2} : \frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}

. This means the line segment connecting your point

P

to the intersection point

B

L_{1}

must be parallel to

L_{2}

. Therefore, the direction ratios of

P B

must be proportional to the direction ratios of

L_{2}

, which are

(2, - 3, 6)

Is it possible for

λ

to be a fraction?

Absolutely! In many JEE problems, the parameter

λ

can be a fraction, a negative number, or even zero. Do not be alarmed if you get a non-integer value. In this specific problem, we found

λ = - 3

, which is a clean integer, but always trust your algebraic steps regardless of the result.

Animated Solution for Mathematics - Three Dimensional Geometry: The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :

Select Answer:

Visualized Solution

Visualizing the Problem

Defining a General Point $B$ on $L_{1}$

Understanding 'Distance Along a Line'

Finding Direction Ratios of $P B$

Setting up the Proportionality

Solving for $λ$

Finding the Coordinates of $B$

Applying the Distance Formula

Final Calculation

Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

The distance of the point $(7, 10, 11)$ from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$ . The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line $L$ along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is:

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

Let the line passing through the points $(- 1, 2, 1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x+2}{3}= rac{y-3}{2}= rac{z-4}{1}$ at the point P. Then the distance of P from the point $Q (4, - 5, 1)$ is

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

Let $P$ be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of $P$ from the line $4 x = 2 y = z$ is

Let $P$ be the point $(10, - 2, - 1)$ and $Q$ be the foot of the perpendicular drawn from the point $R (1, 7, 6)$ on the line passing through the points $(2, - 5, 11)$ and $(- 6, 7, - 5)$ . Then the length of the line segment $P Q$ is equal to ________

The distance of the point having position vector $- \hat{i} + 2 \hat{j} + 6 \hat{k}$ from the straight line passing through the point $(2, 3, - 4)$ and parallel to the vector, $6 \hat{i} + 3 \hat{j} - 4 \hat{k}$ is :

Select Answer:

Visualized Solution

Visualizing the Problem

Defining a General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Understanding 'Distance Along a Line'

Finding Direction Ratios of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PB

Setting up the Proportionality

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Applying the Distance Formula

Final Calculation

Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (7,10,11) from the line 1x−4​=0y−4​=3z−2​ along the line 2x−9​=3y−13​=6z−17​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the line 2x−2​=3y−6​=4z−3​ from the point (1,4,0) along the line 1x​=2y−2​=3z+3​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of line 3y−2z−1=0=3x−z+4 from the point (2,−1,6) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The square of the distance of the point (715​,732​,7) from the line 3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^​+7k^ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The perpendicular distance, of the line 2x−1​=−1y+2​=2z+3​ from the point P(2,−10,1), is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be the point of intersection of the lines 1x−2​=5y−4​=1z−2​ and 2x−3​=3y−2​=2z−3​. Then, the shortest distance of P from the line 4x=2y=z is

Defining a General Point $B$ on $L_{1}$

Finding Direction Ratios of $P B$

Solving for $λ$

Finding the Coordinates of $B$

The distance of the point $(7, 10, 11)$ from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is:

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

Let $P$ be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of $P$ from the line $4 x = 2 y = z$ is