JEE Main 2025 April

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: Let the area of the triangle formed by the lines $x + 2 = y - 1 = z, \frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x - 3}{- 3} = \frac{y - 3}{5} = \frac{z - 2}{1}$ be $A$ . Then $A^{2}$ is equal to ________

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle in 3D Space

Given lines:
Line 1 ( $L_{1}$ ): $x + 2 = y - 1 = z$
Line 2 ( $L_{2}$ ): $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$
Line 3 ( $L_{3}$ ): $\frac{x - 3}{- 3} = \frac{y - 3}{5} = \frac{z - 2}{1}$

Symmetric Form of Line $L_{1}$

Rewrite $L_{1}$ in symmetric form:
$\frac{x + 2}{1} = \frac{y - 1}{1} = \frac{z}{1} = λ$
General point on $L_{1}$ : $(λ - 2, λ + 1, λ)$

Finding Vertex $A$ : Intersection of $L_{1}$ and $L_{2}$

For $L_{2}$ : $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1} = μ$
General point on $L_{2}$ : $(5 μ + 3, - μ, μ + 1)$
Equating $z$ coordinates: $λ = μ + 1$
Equating $y$ coordinates: $λ + 1 = - μ$

Solving for Vertex $A$

Substitute $λ = μ + 1$ into $λ + 1 = - μ$ :
$(μ + 1) + 1 = - μ \Rightarrow 2 μ = - 2 \Rightarrow μ = - 1$
Then $λ = - 1 + 1 = 0$
Vertex $A = (0 - 2, 0 + 1, 0) = (- 2, 1, 0)$

Finding Vertex $C$ : Intersection of $L_{1}$ and $L_{3}$

For $L_{3}$ : $\frac{x - 3}{- 3} = \frac{y - 3}{5} = \frac{z - 2}{1} = γ$
General point on $L_{3}$ : $(- 3 γ + 3, 5 γ + 3, γ + 2)$
Equating $z$ coordinates: $λ = γ + 2$
Equating $y$ coordinates: $λ + 1 = 5 γ + 3$

Solving for Vertex $C$

Substitute $λ = γ + 2$ into $y$ equation:
$(γ + 2) + 1 = 5 γ + 3 \Rightarrow γ + 3 = 5 γ + 3 \Rightarrow 4 γ = 0 \Rightarrow γ = 0$
Then $λ = 0 + 2 = 2$
Vertex $C = (2 - 2, 2 + 1, 2) = (0, 3, 2)$

Finding Vertex $B$ : Intersection of $L_{2}$ and $L_{3}$

Similarly, solving for $L_{2} \cap L_{3}$ :
Vertex $B = (3, 0, 1)$

Defining Vectors $A B$ and $A C$

Vector $A B = (3 - (- 2)) \hat{i} + (0 - 1) \hat{j} + (1 - 0) \hat{k} = 5 \hat{i} - \hat{j} + \hat{k}$
Vector $A C = (0 - (- 2)) \hat{i} + (3 - 1) \hat{j} + (2 - 0) \hat{k} = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}$

The Vector Area Formula

Area of triangle $A = \frac{1}{2} ∣ A B \times A C ∣$
We need $A^{2} = \frac{1}{4} ∣ A B \times A C ∣^{2}$

Computing the Cross Product $A B \times A C$

$A B \times A C = \hat{i} 52 \hat{j} - 1 2 \hat{k} 12$
$= \hat{i} (- 2 - 2) - \hat{j} (10 - 2) + \hat{k} (10 + 2)$
$= - 4 \hat{i} - 8 \hat{j} + 12 \hat{k}$

Calculating the Magnitude Squared

$∣ A B \times A C ∣^{2} = (- 4)^{2} + (- 8)^{2} + 1 2^{2}$
$= 16 + 64 + 144$
$= 224$

Final Result: $A^{2}$

$A^{2} = \frac{1}{4} \times 224$
$A^{2} = 56$
Key Takeaway: Intersection points of lines in 3D form the vertices of the triangle. The area is half the magnitude of the cross product of two adjacent side vectors.

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

To find the area of the triangle formed by the intersection of three lines, we must first determine the coordinates of the vertices

A

B

, and

C

. We represent each line parametrically to identify the points of intersection.

For line

L_{1}

(

x + 2 = y - 1 = z = λ

), the general point is:

P_{1} = (λ - 2, λ + 1, λ)

For line

L_{2}

(

x = 5 μ + 3, y = - μ, z = μ + 1

), the general point is:

P_{2} = (5 μ + 3, - μ, μ + 1)

For line

L_{3}

(

x = - 3 γ + 3, y = 5 γ + 3, z = γ + 2

), the general point is:

P_{3} = (- 3 γ + 3, 5 γ + 3, γ + 2)

Finding the Vertices

Vertex

A

is the intersection of

L_{1}

and

L_{2}

. Equating their coordinates:

λ = μ + 1 and λ + 1 = - μ

Solving this system yields

μ = - 1

and

λ = 0

. Substituting

λ = 0

into the expression for

L_{1}

, we obtain:

A = (- 2, 1, 0)

By repeating this process for the intersections of

L_{1}

with

L_{3}

(Vertex

C

) and

L_{2}

with

L_{3}

(Vertex

B

), we find:

C = (0, 3, 2)

B = (3, 0, 1)

The Vector Calculation

With the vertices

A (- 2, 1, 0)

B (3, 0, 1)

, and

C (0, 3, 2)

identified, we define the vectors forming two sides of the triangle:

A B = (3 - (- 2)) \hat{i} + (0 - 1) \hat{j} + (1 - 0) \hat{k} = 5 \hat{i} - \hat{j} + \hat{k}

A C = (0 - (- 2)) \hat{i} + (3 - 1) \hat{j} + (2 - 0) \hat{k} = 2 \hat{i} + 2 \hat{j} + 2 \hat{k}

The area

S

of the triangle is given by

S = \frac{1}{2} ∣ A B \times A C ∣

. Consequently, the square of the area is:

S^{2} = \frac{1}{4} ∣ A B \times A C ∣^{2}

Final Calculation

We compute the cross product

V = A B \times A C

V = \hat{i} 52 \hat{j} - 1 2 \hat{k} 12 = \hat{i} (- 2 - 2) - \hat{j} (10 - 2) + \hat{k} (10 + 2) = - 4 \hat{i} - 8 \hat{j} + 12 \hat{k}

Next, we calculate the magnitude squared of this vector:

∣ V ∣^{2} = (- 4)^{2} + (- 8)^{2} + (12)^{2} = 16 + 64 + 144 = 224

Finally, we determine the square of the area:

S^{2} = \frac{1}{4} \times 224 = 56

The square of the area of the triangle is 56.

Common Traps & FAQs

Why can't we just use the distance formula to find the side lengths and then use Heron's formula?

You absolutely could! You could find the length of each side using the distance formula and then apply Heron's formula. However, in 3D space, the vector cross product is significantly more elegant and less prone to algebraic errors involving square roots. It is the standard, most efficient tool for this type of problem.

How do I know if the lines intersect? What if they are skew?

The problem explicitly states that the lines form a triangle. In 3D geometry, lines are often skew, but here they are constrained to intersect. By setting the parametric coordinates equal, you are essentially solving for the unique point where the paths cross. If they didn't intersect, you wouldn't be able to find a consistent value for the parameters.

Why do we use parameters like

λ, μ, γ

Think of these parameters as 'position' markers along the line. They allow us to describe an infinite number of points on a line using a single variable. This is the key to unlocking the intersection point because it reduces a 3D coordinate problem into a simple system of linear equations.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the area of the triangle formed by the lines $x + 2 = y - 1 = z, \frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x - 3}{- 3} = \frac{y - 3}{5} = \frac{z - 2}{1}$ be $A$ . Then $A^{2}$ is equal to ________

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle in 3D Space

Symmetric Form of Line $L_{1}$

Finding Vertex $A$ : Intersection of $L_{1}$ and $L_{2}$

Solving for Vertex $A$

Finding Vertex $C$ : Intersection of $L_{1}$ and $L_{3}$

Solving for Vertex $C$

Finding Vertex $B$ : Intersection of $L_{2}$ and $L_{3}$

Defining Vectors $A B$ and $A C$

The Vector Area Formula

Computing the Cross Product $A B \times A C$

Calculating the Magnitude Squared

Final Result: $A^{2}$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Vertices

The Vector Calculation

Final Calculation

Similar Questions

Consider the lines $L_{1} : x - 1 = y - 2 = z$ and $L_{2} : x - 2 = y = z - 1$ . Let the feet of the perpendiculars from the point $P (5, 1, - 3)$ on the lines $L_{1}$ and $L_{2}$ be $Q$ and $R$ respectively. If the area of the triangle $P QR$ is $A$ , then $4 A^{2}$ is equal to :

Let in a $△ A B C$ the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x-6}{3}= rac{y-7}{2}= rac{z-7}{-2}.$ Then the area (in sq. units) of $△ A B C$ 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 vertices B and C of a $Δ A B C$ lie on the line, $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A (1, - 1, 2)$ , is :-

Let a line pass through two distinct points $P (- 2, - 1, 3)$ and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ If the distance of the point Q from the point $R (1, 3, 3)$ is 5, then the square of the area of $△ P QR$ is equal to:

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

Let P be the image of the point $Q (7, - 2, 5)$ in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R (5, p, q)$ be a point on L. Then the square of the area of $△ P QR$ is

Let a line $L$ pass through the point $P (2, 3, 1)$ and be parallel to the line $x + 3 y - 2 z - 2 = 0 = x - y + 2 z$ . If the distance of $L$ from the point $(5, 3, 8)$ is $α$ , then $3 α^{2}$ 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

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle in 3D Space

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

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

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

Defining Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

The Vector Area Formula

Computing the Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB×AC

Calculating the Magnitude Squared

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Vertices

The Vector Calculation

Final Calculation

Similar Questions

.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; } Let P be the image of the point Q(7,−2,5) in the line L:2x−1​=3y+1​=4z​ and R(5,p,q) be a point on L. Then the square of the area of △PQR is

Symmetric Form of Line $L_{1}$

Solving for Vertex $A$

Solving for Vertex $C$

Defining Vectors $A B$ and $A C$

Computing the Cross Product $A B \times A C$

Final Result: $A^{2}$

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 image of the point $Q (7, - 2, 5)$ in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R (5, p, q)$ be a point on L. Then the square of the area of $△ P QR$ is