Why do we use different parameters $\lambda$ and $\mu$ for the two lines?

Because the lines are independent entities in 3D space. The point of intersection is a specific location where the lines meet, but the parameters $\lambda$ and $\mu$ represent the position along each line independently. If you used the same parameter, you would be forcing the lines to meet at the same relative position, which is not necessarily true.

JEE Main 2024 (29 Jan Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let $P QR$ be a triangle with $R (- 1, 4, 2)$ . Suppose $M (2, 1, 2)$ is the mid point of $P Q$ . The distance of the centroid of $△ P QR$ from the point of intersection of the line $\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}$ is

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the triangle

P QR

in 3D space. You are given a vertex

R

and the midpoint

M

of the opposite side

P Q

. The segment

R M

is a median. The centroid

G

is not just any point; it is the geometric center that divides the median

R M

in a specific ratio. Can you recall that ratio?

Hint 2

The Arsenal: To find the centroid

G

, use the section formula. If a point divides a line segment joining

(x_{1}, y_{1}, z_{1})

and

(x_{2}, y_{2}, z_{2})

in the ratio

m : n

, the coordinates are

(\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n}, \frac{m z _{2} + n z _{1}}{m + n})

. For the lines, remember that the intersection point must satisfy the equations of both lines simultaneously.

Hint 3

The Execution: Parametrize the lines

L_{1}

and

L_{2}

using parameters

λ

and

μ

respectively. For

L_{1}

, the general point is

(2, 2 λ, - λ - 3)

. For

L_{2}

, it is

(μ + 1, - 3 μ - 3, μ - 1)

. Equate the

x

-coordinates to find

μ

, then find the intersection point

A

. Finally, calculate the distance

G A

using the distance formula.

Visualized Solution

Geometry of $△ P QR$

Given vertex $R (- 1, 4, 2)$
Given midpoint of $P Q$ as $M (2, 1, 2)$
The line segment $R M$ is the median of $△ P QR$

The Centroid Property

Centroid $G$ divides the median $R M$ in the ratio $2 : 1$ from vertex $R$
Using Section Formula: $G = \frac{2 M + 1 R}{2 + 1}$

Setup for Centroid $G$

$x_{G} = \frac{2 ( 2 ) + 1 ( - 1 )}{3}$
$y_{G} = \frac{2 ( 1 ) + 1 ( 4 )}{3}$
$z_{G} = \frac{2 ( 2 ) + 1 ( 2 )}{3}$

Coordinates of $G$

$x_{G} = \frac{3}{3} = 1$
$y_{G} = \frac{6}{3} = 2$
$z_{G} = \frac{6}{3} = 2$
Centroid $G = (1, 2, 2)$

Parametric Form of $L_{1}$

Line $L_{1} : \frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1} = λ$
General point on $L_{1}$ : $(2, 2 λ, - λ - 3)$

Parametric Form of $L_{2}$

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

Intersection Setup

Equating $x$ -coordinates of general points:
$2 = μ + 1$
$μ = 1$

Intersection Point $A$

Substitute $μ = 1$ into general point of $L_{2}$ :
$y = - 3 (1) - 3 = - 6$
$z = 1 - 1 = 0$
Intersection point $A = (2, - 6, 0)$

Distance Formula Setup

Points: $G (1, 2, 2)$ and $A (2, - 6, 0)$
Distance $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$

Substituting Coordinates

$d = (2 - 1)^{2} + (- 6 - 2)^{2} + (0 - 2)^{2}$
$d = (1)^{2} + (- 8)^{2} + (- 2)^{2}$

Final Calculation

$d = 1 + 64 + 4$
$d = 69$

00:00 / 00:00

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Geometry of the Triangle

Imagine a triangle

P QR

in a 3D coordinate system. We are given the vertex

R (- 1, 4, 2)

and the midpoint

M (2, 1, 2)

of the side

P Q

. The line segment

R M

represents the median of the triangle.

The centroid

G

is the point where all three medians intersect. A fundamental property of the centroid is that it divides each median in a

2 : 1

ratio, measured from the vertex.

Using the section formula, we determine the coordinates of

G

by dividing the segment

R M

in the ratio

2 : 1

G = \frac{2 M + 1 R}{2 + 1}

Substituting the given coordinates into the formula:

G = \frac{2 ( 2 , 1 , 2 ) + ( - 1 , 4 , 2 )}{3} = \frac{( 4 - 1 , 2 + 4 , 4 + 2 )}{3} = \frac{( 3 , 6 , 6 )}{3} = (1, 2, 2)

Thus, the coordinates of the centroid are

G (1, 2, 2)

Parametrizing the Lines

Next, we consider the two lines

L_{1}

and

L_{2}

. To find their intersection point

A

, we express both lines in parametric form.

For

L_{1} : \frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1} = λ

, the general point is

(2, 2 λ, - λ - 3)

For

L_{2} : \frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1} = μ

, the general point is

(μ + 1, - 3 μ - 3, μ - 1)

At the point of intersection

A

, these coordinates must be identical. Equating the

x

-coordinates:

2 = μ + 1 \Rightarrow μ = 1

Substituting

μ = 1

into the general point of

L_{2}

, we find the intersection point:

A = (1 + 1, - 3 (1) - 3, 1 - 1) = (2, - 6, 0)

Calculating the Final Distance

We now have the two points: the centroid

G (1, 2, 2)

and the intersection point

A (2, - 6, 0)

. We calculate the distance

d

between them using the 3D distance formula:

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Substituting the values:

d = (2 - 1)^{2} + (- 6 - 2)^{2} + (0 - 2)^{2}

d = 1^{2} + (- 8)^{2} + (- 2)^{2} = 1 + 64 + 4 = 69

The final distance between the centroid

G

and the intersection point

A

is $69$ .

Common Traps & FAQs

Why do we use different parameters

λ

and

μ

for the two lines?

Because the lines are independent entities in 3D space. The point of intersection is a specific location where the lines meet, but the parameters

λ

and

μ

represent the position along each line independently. If you used the same parameter, you would be forcing the lines to meet at the same relative position, which is not necessarily true.

Can we just use the centroid formula

G = (P + Q + R) /3

That is a valid formula, but we don't have the coordinates of

P

and

Q

individually! We only have the midpoint

M

P Q

. However, notice that

M = (P + Q) /2

. So,

(P + Q) = 2 M

. Substituting this into the centroid formula gives

G = (2 M + R) /3

. This is exactly what we used!

How do I know if the lines actually intersect?

In this problem, we are told they intersect. If you were not sure, you would equate the

x, y,

and

z

coordinates and solve for

λ

and

μ

using two equations (e.g.,

x

and

y

), then check if those values satisfy the third equation (

z

). If they do, the lines intersect. If not, they are skew lines.

Select Answer:

Visualized Solution

Geometry of $△ P QR$

The Centroid Property

Setup for Centroid $G$

Coordinates of $G$

Parametric Form of $L_{1}$

Parametric Form of $L_{2}$

Intersection Setup

Intersection Point $A$

Distance Formula Setup

Substituting Coordinates

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Geometry of the Triangle

Parametrizing the Lines

Calculating the Final Distance

Similar Questions

Let $P$ and $Q$ be the points on the line $\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ which are at a distance of 6 units from the point $R (1, 2, 3)$ . If the centroid of the triangle $P QR$ is $(α, β, γ)$ , then $α^{2} + β^{2} + γ^{2}$ 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

Let the vertices $Q$ and $R$ of the triangle $P QR$ lie on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ , $QR = 5$ and the coordinates of the point $P$ be $(0, 2, 3)$ . If the area of the triangle $P QR$ is $\frac{m}{n}$ then :

Let $Q$ and $R$ be two points on the line $\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}$ at a distance $26$ from the point $P (4, 2, 7)$ . Then the square of the area of the triangle $P QR$ is ______.

Let P be the foot of the perpendicular from the point $Q (10, - 3, - 1)$ on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ Then the area of the right angled triangle PQR, where R is the point $(3, - 2, 1)$ , is

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ , $λ \in R,$ intersect the line L at Q. Then $2 (P Q)^{2}$ is equal to:

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

If two distinct point $Q, R$ lie on the line of intersection of the planes $- x + 2 y - z = 0$ and $3 x - 5 y + 2 z = 0$ and $P Q = P R = 18$ where the point $P$ is $(1, - 2, 3)$ , then the area of the triangle $P QR$ 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:

Select Answer:

Visualized Solution

Geometry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PQR

The Centroid Property

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

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

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

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

Intersection Setup

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

Distance Formula Setup

Substituting Coordinates

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Geometry of the Triangle

Parametrizing the Lines

Calculating the Final Distance

Similar Questions

.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; } Let Q and R be two points on the line 2x+1​=3y+2​=2z−1​ at a distance 26​ from the point P(4,2,7). Then the square of the area of the triangle PQR 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:

Geometry of $△ P QR$

Setup for Centroid $G$

Coordinates of $G$

Parametric Form of $L_{1}$

Parametric Form of $L_{2}$

Intersection Point $A$

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 $Q$ and $R$ be two points on the line $\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}$ at a distance $26$ from the point $P (4, 2, 7)$ . Then the square of the area of the triangle $P QR$ 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: