Why do we use $\lambda$ and $\mu$ instead of just one variable?

Great question! Using the same parameter would force the lines to intersect at a specific relative position, which is rarely true. By using $\lambda$ for the first line and $\mu$ for the second, we allow each line to be independent. The intersection is the specific pair of $(\lambda, \mu)$ that makes the coordinates match.

JEE Main 2024 (05 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let $d$ be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point $(7, 8, 9)$ . Then $d^{2} + 6$ is equal to :

Q: Why calculate $d^2$ instead of $d$?

Efficiency! The question asks for $d^2 + 6$. Calculating $d$ would involve a square root, which we would just square again anyway. By calculating $d^2$ directly, we save time and avoid potential errors with radicals.

Select Answer:

Progressive Hints

Hint 1

Visualize the lines as infinite paths in 3D space. To find a point on a line, you need a 'map'—a parameter. Can you express the coordinates

(x, y, z)

of any point on

L_{1}

using a single variable

λ

by setting the ratios equal to

λ

Hint 2

When two lines intersect, they share a common point. This means at that specific point, the

x, y,

and

z

coordinates from both lines must be identical. You need two different parameters, say

λ

and

μ

, to describe points on

L_{1}

and

L_{2}

independently, because the intersection doesn't necessarily happen at the same 'time' or 'position' along each line.

Hint 3

Set up your system: equate the

x, y,

and

z

expressions from your parameterizations. You will get a system like

3 λ - 6 = 4 μ + 7

and

2 λ = 3 μ + 9

. Solve this system for

λ

and

μ

to find the unique intersection point

P

. Finally, use the distance formula between

P

and

(7, 8, 9)

Visualized Solution

Visualizing the Problem

Given lines: $L_{1} : \frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$
Given lines: $L_{2} : \frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$
Target: Find distance $d$ of intersection point from $(7, 8, 9)$

Parametrizing Line $L_{1}$

Let $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1} = λ$
General point on $L_{1}$ :
$(x, y, z) = (3 λ - 6, 2 λ, λ - 1)$

Parametrizing Line $L_{2}$

Let $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2} = μ$
General point on $L_{2}$ :
$(x, y, z) = (4 μ + 7, 3 μ + 9, 2 μ + 4)$

Equating Coordinates at Intersection

At intersection point $P$ , coordinates must be equal.
$x$ -coordinate: $3 λ - 6 = 4 μ + 7 ⟹ 3 λ - 4 μ = 13$
$y$ -coordinate: $2 λ = 3 μ + 9 ⟹ 2 λ - 3 μ = 9$

Solving for $λ$ and $μ$

Eq 1: $3 λ - 4 μ = 13$
Eq 2: $2 λ - 3 μ = 9$
Multiply Eq 1 by $2$ and Eq 2 by $3$ :
$6 λ - 8 μ = 26$
$6 λ - 9 μ = 27$
Subtracting gives: $μ = - 1$
Substituting $μ = - 1$ in Eq 2 gives: $λ = 3$

Finding the Intersection Point $P$

Substitute $λ = 3$ into $L_{1}$ general point:
$x = 3 (3) - 6 = 3$
$y = 2 (3) = 6$
$z = 3 - 1 = 2$
Intersection Point $P = (3, 6, 2)$

Setting up the Distance Calculation

Intersection Point $P = (3, 6, 2)$
Given Point $Q = (7, 8, 9)$
We need the distance $d$ between $P$ and $Q$ .

Applying the Distance Formula

Distance formula: $d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Substitute $P (3, 6, 2)$ and $Q (7, 8, 9)$ :
$d^{2} = (7 - 3)^{2} + (8 - 6)^{2} + (9 - 2)^{2}$

Calculating $d^{2}$

$d^{2} = (4)^{2} + (2)^{2} + (7)^{2}$
$d^{2} = 16 + 4 + 49$
$d^{2} = 69$

Final Answer: $d^{2} + 6$

We need to find the value of $d^{2} + 6$ .
Substitute $d^{2} = 69$ :
$d^{2} + 6 = 69 + 6 = 75$
Final Answer: 75

00:00 / 00:00

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

To find the intersection point of two lines in 3D space, we represent each line using a unique parameter. This allows us to define any arbitrary point on either line as a function of that parameter.

For the first line,

L_{1}

, given by:

\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1} = λ

We can express any point on

L_{1}

(3 λ - 6, 2 λ, λ - 1)

. Here,

λ

acts as a scalar coordinate along the line.

For the second line,

L_{2}

, given by:

\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2} = μ

We express any point on

L_{2}

(4 μ + 7, 3 μ + 9, 2 μ + 4)

using a different parameter,

μ

The Collision Course

At the point of intersection

P

, the coordinates derived from

L_{1}

must be identical to those derived from

L_{2}

. This yields a system of linear equations:

3 λ - 6 = 4 μ + 7 \Rightarrow 3 λ - 4 μ = 13

2 λ = 3 μ + 9 \Rightarrow 2 λ - 3 μ = 9

To solve this system, we multiply the first equation by

2

and the second by

3

6 λ - 8 μ = 26

6 λ - 9 μ = 27

Subtracting the second equation from the first eliminates

λ

, resulting in

μ = - 1

. Substituting

μ = - 1

back into the equation

2 λ - 3 (- 1) = 9

, we find

2 λ = 6

, which gives

λ = 3

Final Calculation

With

λ = 3

, we determine the coordinates of the intersection point

P

by substituting

λ

into the expression for

L_{1}

x = 3 (3) - 6 = 3

y = 2 (3) = 6

z = 3 - 1 = 2

Thus, the intersection point is

P (3, 6, 2)

. We now calculate the squared distance

d^{2}

between

P (3, 6, 2)

and the point

Q (7, 8, 9)

using the distance formula:

d^{2} = (7 - 3)^{2} + (8 - 6)^{2} + (9 - 2)^{2}

d^{2} = 4^{2} + 2^{2} + 7^{2} = 16 + 4 + 49 = 69

The problem asks for the value of

d^{2} + 6

. Therefore, the final result is:

69 + 6 = 75

Common Traps & FAQs

Why do we use

λ

and

μ

instead of just one variable?

Great question! Using the same parameter would force the lines to intersect at a specific relative position, which is rarely true. By using

λ

for the first line and

μ

for the second, we allow each line to be independent. The intersection is the specific pair of

(λ, μ)

that makes the coordinates match.

What if the system of equations has no solution?

If the system is inconsistent, it means the lines are skew—they never meet in 3D space. In JEE problems, if you are asked to find the intersection point, the lines are guaranteed to intersect, but always check for consistency to be sure!

Why calculate

d^{2}

instead of

d

Efficiency! The question asks for

d^{2} + 6

. Calculating

d

would involve a square root, which we would just square again anyway. By calculating

d^{2}

directly, we save time and avoid potential errors with radicals.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $d$ be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point $(7, 8, 9)$ . Then $d^{2} + 6$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Problem

Parametrizing Line $L_{1}$

Parametrizing Line $L_{2}$

Equating Coordinates at Intersection

Solving for $λ$ and $μ$

Finding the Intersection Point $P$

Setting up the Distance Calculation

Applying the Distance Formula

Calculating $d^{2}$

Final Answer: $d^{2} + 6$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Collision Course

Final Calculation

Similar Questions

Let $L$ be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$ , whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line $L$ is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to :

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 :

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

If the distances of the point $(1, 2, a)$ from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and $L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to

The lines $\frac{x - 2}{2} = \frac{y}{- 2} = \frac{z - 7}{16}$ and $\frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1}$ intersect at the point $P$ . If the distance of $P$ from the line $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1}$ is $l$ , then $14 l^{2}$ is equal to

Let $Q (a, b, c)$ be the image of the point $P (3, 2, 1)$ in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ . Then the distance of $Q$ from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is

Let the point, on the line passing through the points $P (1, - 2, 3)$ and $Q (5, - 4, 7)$ , farther from the origin and at distance of 9 units from the point $P$ , be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to :

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

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:

Let the line $L$ be the projection of the line $\frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 4}{2}$ in the plane $x - 2 y - z = 3$ . If $d$ is the distance of the point $(0, 0, 6)$ from $L$ , then $d^{2}$ is equal to .

Select Answer:

Visualized Solution

Visualizing the Problem

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

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

Equating Coordinates at Intersection

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

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Setting up the Distance Calculation

Applying the Distance Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2+6

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Collision Course

Final Calculation

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,−2,11) from the line 1x−6​=0y−4​=3z−8​ along the line 2x−5​=−3y−1​=6z−5​, is :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Q(a,b,c) be the image of the point P(3,2,1) in the line 1x−1​=2y​=1z−1​. Then the distance of Q from the line 3x−9​=2y−9​=−2z−5​ 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

.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; } Let the line L be the projection of the line 2x−1​=1y−3​=2z−4​ in the plane x−2y−z=3. If d is the distance of the point (0,0,6) from L, then d2 is equal to .

Parametrizing Line $L_{1}$

Parametrizing Line $L_{2}$

Solving for $λ$ and $μ$

Finding the Intersection Point $P$

Calculating $d^{2}$

Final Answer: $d^{2} + 6$

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 :

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

Let $Q (a, b, c)$ be the image of the point $P (3, 2, 1)$ in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ . Then the distance of $Q$ from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ 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

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:

Let the line $L$ be the projection of the line $\frac{x - 1}{2} = \frac{y - 3}{1} = \frac{z - 4}{2}$ in the plane $x - 2 y - z = 3$ . If $d$ is the distance of the point $(0, 0, 6)$ from $L$ , then $d^{2}$ is equal to .