Three lines are given by r⃗ = λî, λ ∈ R; r⃗ = μ(î + ĵ), μ ∈ R and r⃗ = ν(î + ĵ + k̂), ν ∈ R. Le

Q: Why do we use the parametric form $\lambda, \mu, \nu$ instead of just solving the equations directly?

Great question! The parametric form is the bridge between the line's direction and the plane's constraint. By setting $\vec{r} = \lambda \hat{i}$, you are essentially saying, 'I am looking for a point that satisfies both the line's direction and the plane's equation.' It turns a 3D geometry problem into a simple algebraic substitution. It is the most elegant way to 'pierce' the plane.

Q: Is there a shortcut to finding the area without the cross product?

While you could use coordinate geometry formulas for the area of a triangle in 3D, the cross product is the 'gold standard' for JEE Advanced. It handles the orientation and magnitude naturally. Trying to use distance formulas or projections often leads to messy square roots and higher chances of calculation errors. Stick to the cross product; it is reliable and mathematically beautiful.

Q: Wait, why do we get such simple fractions like $1/2$ and $1/3$? Did I miss something?

You didn't miss anything! The problem is designed with symmetry in mind. The plane $x + y + z = 1$ is a very special plane in 3D space—it's the plane that intercepts the axes at $(1,0,0), (0,1,0),$ and $(0,0,1)$. The lines are also symmetric. When you see clean fractions in a JEE problem, it is usually a sign that you are on the right track. Trust the math!

Question

Three lines are given by $\vec{r} = \lambda \hat{i}, \lambda \in \mathbb{R}$; $\vec{r} = \mu(\hat{i} + \hat{j}), \mu \in \mathbb{R}$ and $\vec{r} = 
u(\hat{i} + \hat{j} + \hat{k}), 
u \in \mathbb{R}$. Let the lines cut the plane $x + y + z = 1$ at the points $A, B$ and $C$ respectively. If the area of the triangle $ABC$ is $\Delta$ then the value of $(6\Delta)^2$ equals

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are visualizing a scene in three-dimensional space. Imagine you are standing at the origin of a coordinate system with three laser pointers shooting beams in specific directions.

The first beam travels along the $x$-axis. The second beam cuts through the $x-y$ plane. The third beam slices diagonally through the heart of the octant.

We are tasked with finding where these beams pierce the plane defined by the equation $x + y + z = 1$. Our ultimate goal is to calculate the area of the triangle formed by these three points of impact.

### Phase 1: The Piercing Points

To find where these lines hit the plane, we use parametric equations. A line is a collection of points; by representing a line as $\vec{r} = \lambda \hat{i}$, we define any point on this line as $(\lambda, 0, 0)$.

When we force this point to satisfy the plane equation $x + y + z = 1$, we find the intersection:
$$\lambda + 0 + 0 = 1 \Rightarrow \lambda = 1$$
Thus, our first point, $A$, is $(1, 0, 0)$.

For the second line, $\vec{r} = \mu(\hat{i} + \hat{j})$, the coordinates are $(\mu, \mu, 0)$. Substituting into the plane equation:
$$\mu + \mu + 0 = 1 \Rightarrow 2\mu = 1 \Rightarrow \mu = \frac{1}{2}$$
Thus, point $B$ is $(\frac{1}{2}, \frac{1}{2}, 0)$.

For the third line, $\vec{r} = 
u(\hat{i} + \hat{j} + \hat{k})$, the coordinates are $(
u, 
u, 
u)$. Substituting into the plane equation:
$$
u + 
u + 
u = 1 \Rightarrow 3
u = 1 \Rightarrow 
u = \frac{1}{3}$$
Thus, point $C$ is $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$.

### Phase 2: The Vector Construction

Now that we have our vertices $A(1, 0, 0)$, $B(\frac{1}{2}, \frac{1}{2}, 0)$, and $C(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$, we define the triangle using two adjacent side vectors, $\vec{AB}$ and $\vec{AC}$.

Calculating $\vec{AB}$ via subtraction of position vectors:
$$\vec{AB} = B - A = \left( \frac{1}{2} - 1 ight)\hat{i} + \left( \frac{1}{2} - 0 ight)\hat{j} + (0 - 0)\hat{k} = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$$

Similarly, for $\vec{AC}$:
$$\vec{AC} = C - A = \left( \frac{1}{3} - 1 ight)\hat{i} + \left( \frac{1}{3} - 0 ight)\hat{j} + \left( \frac{1}{3} - 0 ight)\hat{k} = -\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{1}{3}\hat{k}$$

### Phase 3: The Engine of the Solution

The area $\Delta$ of a triangle in 3D space is given by the formula $\Delta = \frac{1}{2} |\vec{AB} 	imes \vec{AC}|$. We set up the determinant for the cross product:

$$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -\frac{1}{2} & \frac{1}{2} & 0 \ -\frac{2}{3} & \frac{1}{3} & \frac{1}{3} \end{vmatrix}$$

Expanding this determinant:
- $\hat{i}$ component: $(\frac{1}{2} \cdot \frac{1}{3}) - 0 = \frac{1}{6}$
- $\hat{j}$ component: $-\left( (-\frac{1}{2} \cdot \frac{1}{3}) - 0 ight) = \frac{1}{6}$
- $\hat{k}$ component: $(-\frac{1}{2} \cdot \frac{1}{3}) - (-\frac{2}{3} \cdot \frac{1}{2}) = -\frac{1}{6} + \frac{2}{6} = \frac{1}{6}$

The resulting vector is $\frac{1}{6}\hat{i} + \frac{1}{6}\hat{j} + \frac{1}{6}\hat{k}$. Its magnitude is:
$$|\vec{AB} 	imes \vec{AC}| = \sqrt{\left(\frac{1}{6}ight)^2 + \left(\frac{1}{6}ight)^2 + \left(\frac{1}{6}ight)^2} = \sqrt{\frac{3}{36}} = \frac{\sqrt{3}}{6}$$

### Final Calculation

The area $\Delta$ is half of this magnitude:
$$\Delta = \frac{1}{2} \cdot \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{12}$$

The problem asks for the value of $(6\Delta)^2$. First, calculate $6\Delta$:
$$6\Delta = 6 \cdot \frac{\sqrt{3}}{12} = \frac{\sqrt{3}}{2}$$

Squaring this result yields:
$$\left( \frac{\sqrt{3}}{2} ight)^2 = \frac{3}{4} = \mathbf{0.75}$$

Enter Numerical Value:

Visualized Solution

The 3D Setup

Intersection Strategy

Finding Point A

Finding Point B

Finding Point C

Area of Triangle Formula

Vectors $A B$ and $A C$

Cross Product Setup

Evaluating Cross Product

Magnitude of Cross Product

Calculating Area $Δ$

Evaluating $(6Δ)^{2}$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

Similar Questions

Suppose the line $\frac{x - 2}{α} = \frac{y - 2}{- 5} = \frac{z + 2}{2}$ lies on the plane $x + 3 y - 2 z + β = 0$ . Then $(α + β)$ is equal to .

For real numbers $α$ and $β \neq = 0$ , if the point of intersection of the straight lines $\frac{x - α}{1} = \frac{y - 1}{2} = \frac{z - 1}{3}$ and $\frac{x - 4}{β} = \frac{y - 6}{3} = \frac{z - 7}{3}$ lies on the plane $x + 2 y - z = 8$ , then $α - β$ is equal to :

If the angle $θ$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2 x - y + λ z + 4 = 0$ is such that $sin θ = \frac{1}{3}$ , then the value of $λ$ is

If the line, $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}$ lies in the plane, $l x + m y - z = 9$ , then $l^{2} + m^{2}$ is equal to :