JEE Main 11 Jan 2019 (Evening)

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Animated Solution for Mathematics - Three Dimensional Geometry: Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$ and $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$ intersect at the point $R$ . The reflection of $R$ in the xy-plane has coordinates:

Q: Why does reflecting in the $xy$-plane only change the $z$-coordinate?

Think of the $xy$-plane as the floor of your room. If you are standing at a point $(x, y, z)$, your $x$ and $y$ coordinates define your position on the floor tiles. Your $z$ coordinate is your height above the floor. If you reflect yourself through the floor, your position on the tiles $(x, y)$ doesn't change, but your height becomes negative—you are now the same distance below the floor.

Select Answer:

Progressive Hints

Hint 1

Think of a line not as a static object, but as a path. If you are a traveler on this line, how would you describe your position at any moment? Introduce a parameter, say

λ

, to represent your distance from a reference point. This transforms the symmetric equation into a 'general point'

(x, y, z)

Hint 2

When two lines intersect, they share a common 'meeting point'

R

. At this point, the coordinates

(x, y, z)

must satisfy the equations of both lines simultaneously. Since we have two lines, use two distinct parameters,

λ

for the first and

μ

for the second. Equate the

x

y

, and

z

components to form a system of linear equations.

Hint 3

You have three equations (from

x

y

, and

z

components) but only two unknowns (

λ

and

μ

). Start by solving the system using the

x

and

y

equations. Once you find

λ

and

μ

, verify them by checking if they satisfy the

z

equation. The intersection point

R

is then found by substituting your parameter back into the line equation.

Visualized Solution

Visualizing the Intersecting Lines

Line $L_{1}$ : $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$
Line $L_{2}$ : $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$
Goal: Find the intersection point $R$ and its reflection in the $x y$ -plane.

Parametric Form of $L_{1}$

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

Parametric Form of $L_{2}$

Let $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4} = μ$
General point on $L_{2}$ : $P_{2} (- 5 + 7 μ, 2 - 6 μ, 3 + 4 μ)$

Equating the $x$ -coordinates

At intersection point $R$ , $P_{1}$ and $P_{2}$ must be the same point.
Equating $x$ : $3 + λ = - 5 + 7 μ$
Rearranging: $λ - 7 μ = - 8$ (Equation 1)

Equating the $y$ -coordinates

Equating $y$ : $- 1 + 3 λ = 2 - 6 μ$
$3 λ + 6 μ = 3$
Dividing by $3$ : $λ + 2 μ = 1$ (Equation 2)

Solving for $μ$

Subtracting (Eq 1) from (Eq 2):
$(λ + 2 μ) - (λ - 7 μ) = 1 - (- 8)$
$9 μ = 9 ⟹ μ = 1$

Finding $λ$

Substitute $μ = 1$ into Equation 2:
$λ + 2 (1) = 1$
$λ = 1 - 2 = - 1$

Calculating Point $R$

Substitute $λ = - 1$ into $P_{1} (3 + λ, - 1 + 3 λ, 6 - λ)$ :
$x = 3 - 1 = 2$
$y = - 1 - 3 = - 4$
$z = 6 - (- 1) = 7$
Intersection point $R = (2, - 4, 7)$

Reflection in $x y$ -plane

The $x y$ -plane acts as a mirror at $z = 0$ .
Reflection of a point $(x, y, z)$ in the $x y$ -plane changes the sign of the $z$ -coordinate.
Formula: $(x, y, z) \to (x, y, - z)$

Finding $R^{'}$

Point $R$ is $(2, - 4, 7)$ .
Apply reflection rule: $z \to - z$
Reflected point $R^{'}$ is $(2, - 4, - 7)$ .

Final Result

The coordinates of the reflection of $R$ in the $x y$ -plane are $(2, - 4, - 7)$ .
Correct Option: (2, -4, -7)

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

In 3D space, a line is defined by its direction and a point it passes through. The symmetric form given,

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

is elegant, but it is static. To find an intersection, we must parameterize the line.

By setting the symmetric equation equal to a parameter

λ

, we describe any point on the line as a function of this single variable:

P_{1} = (3 + λ, - 1 + 3 λ, 6 - λ)

This serves as our coordinate on

L_{1}

. As

λ

changes, we traverse the line. We apply the same logic to

L_{2}

using a distinct parameter

μ

P_{2} = (- 5 + 7 μ, 2 - 6 μ, 3 + 4 μ)

The Collision

If the lines intersect at a point

R

, then at that specific location, the coordinates of

P_{1}

and

P_{2}

must be identical. We equate the components to find the intersection:

For the

x

-coordinate:

3 + λ = - 5 + 7 μ \Rightarrow λ - 7 μ = - 8 (Equation 1)

For the

y

-coordinate:

- 1 + 3 λ = 2 - 6 μ \Rightarrow 3 λ + 6 μ = 3

Simplifying the

y

-equation by dividing by

3

, we obtain:

λ + 2 μ = 1 (Equation 2)

Solving the System

We now have a clean system of two linear equations:

λ - 7 μ = - 8

λ + 2 μ = 1

Subtracting the first equation from the second, the

λ

terms vanish, leaving us with

9 μ = 9

, which yields

μ = 1

. Substituting this back into Equation 2, we find

λ = 1 - 2 (1) = - 1

Finding the Intersection Point

With

λ = - 1

, we determine the intersection point

R

by substituting the parameter back into our expression for

P_{1}

x = 3 + (- 1) = 2

y = - 1 + 3 (- 1) = - 4

z = 6 - (- 1) = 7

Thus, the point of intersection is $R = (2, - 4, 7)$ .

The Reflection

We must now find the reflection of

R

in the

x y

-plane. Imagine the

x y

-plane as a mirror at

z = 0

Reflecting a point across this plane preserves the

(x, y)

coordinates while inverting the depth

(z)

. Thus, the transformation is defined as

(x, y, z) \to (x, y, - z)

Applying this to

R (2, - 4, 7)

, we obtain the final reflected point:

$R^{'} = (2, - 4, - 7)$

Common Traps & FAQs

Why do we need two different parameters,

λ

and

μ

? Can't we just use

λ

for both?

That is a fantastic question! If you use the same parameter for both lines, you are essentially forcing the lines to 'arrive' at the intersection point at the same 'time' or 'distance' from their respective origins. But lines are infinite; they don't have a synchronized clock. By using

λ

and

μ

, we allow each line to be independent. We are looking for a geometric intersection, not a synchronized collision.

What if the system of equations has no solution? Does that mean the lines are parallel?

Not necessarily! If the system is inconsistent, it means the lines are 'skew'—they are neither parallel nor intersecting. They pass each other like ships in the night in 3D space. If they were parallel, the direction ratios would be proportional, which is a different check entirely.

Why does reflecting in the

x y

-plane only change the

z

-coordinate?

Think of the

x y

-plane as the floor of your room. If you are standing at a point

(x, y, z)

, your

x

and

y

coordinates define your position on the floor tiles. Your

z

coordinate is your height above the floor. If you reflect yourself through the floor, your position on the tiles

(x, y)

doesn't change, but your height becomes negative—you are now the same distance below the floor.

Animated Solution for Mathematics - Three Dimensional Geometry: Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$ and $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$ intersect at the point $R$ . The reflection of $R$ in the xy-plane has coordinates:

Select Answer:

Visualized Solution

Visualizing the Intersecting Lines

Parametric Form of $L_{1}$

Parametric Form of $L_{2}$

Equating the $x$ -coordinates

Equating the $y$ -coordinates

Solving for $μ$

Finding $λ$

Calculating Point $R$

Reflection in $x y$ -plane

Finding $R^{'}$

Final Result

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Collision

Solving the System

Finding the Intersection Point

The Reflection

Similar Questions

Consider a line $L$ passing through the points $P (1, 2, 1)$ and $Q (2, 1, - 1)$ . If the mirror image of the point $A (2, 2, 2)$ in the line $L$ is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to

Let $(α, β, γ)$ be mirror image of the point $(2, 3, 5)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to

If the mirror image of the point $P (3, 4, 9)$ in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is :

Statement-1: The point $A (1, 0, 7)$ is the mirror image of the point $B (1, 6, 3)$ in the line: $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Statement-2: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A (1, 0, 7)$ and $B (1, 6, 3)$ .

Let a line passing through the point $(4, 1, 0)$ intersect the line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ at the point $A (α, β, γ)$ and the line $L_{2} : x - 6 = y = - z + 4$ at the point $B (a, b, c)$ . Then $1 α a 0 β b 1 γ c$ is equal to

The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is:

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is $P$ , then $a + b$ is equal to .

Select Answer:

Visualized Solution

Visualizing the Intersecting Lines

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​

Equating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinates

Equating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinates

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

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

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

Reflection in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xy-plane

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

Final Result

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Collision

Solving the System

Finding the Intersection Point

The Reflection

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider a line L passing through the points P(1,2,1) and Q(2,1,−1). If the mirror image of the point A(2,2,2) in the line L is (α,β,γ), then α+β+6γ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (α,β,γ) be mirror image of the point (2,3,5) in the line 2x−1​=3y−2​=4z−3​. Then 2α+3β+4γ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mirror image of the point P(3,4,9) in the line 3x−1​=2y+1​=1z−2​ is (α,β,γ), then 14(α+β+γ) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the line passing through (−4,3,1), parallel to the plane x+2y−z−5=0 and intersecting the line −3x+1​=2y−3​=−1z−2​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point P(a,2,a) in the line 2x​=1y+a​=1z​ is Q and the image of Q in the line 2x−2b​=1y−a​=−5z+2b​ is P, then a+b is equal to .

Parametric Form of $L_{1}$

Parametric Form of $L_{2}$

Equating the $x$ -coordinates

Equating the $y$ -coordinates

Solving for $μ$

Finding $λ$

Calculating Point $R$

Reflection in $x y$ -plane

Finding $R^{'}$

Consider a line $L$ passing through the points $P (1, 2, 1)$ and $Q (2, 1, - 1)$ . If the mirror image of the point $A (2, 2, 2)$ in the line $L$ is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to

Let $(α, β, γ)$ be mirror image of the point $(2, 3, 5)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to

If the mirror image of the point $P (3, 4, 9)$ in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is :

The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is:

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is $P$ , then $a + b$ is equal to .