Consider the lines L1 : x - 1 = y - 2 = z and L2 : x - 2 = y = z - 1. Let the feet of the perpendicu

Q: Wait, why do we use the dot product to find the foot of the perpendicular? Can't we just use the distance formula?

You *could* minimize the distance squared, but that is just calculus in disguise! The dot product is the elegant, geometric shortcut. It directly enforces the condition that the vector from $P$ to the line is orthogonal to the line's direction vector. It is faster and less prone to algebraic errors.

Q: Why are the direction vectors of both lines $(1, 1, 1)$? Is that a coincidence?

It is not a coincidence, it is a property of the lines given! Look at the symmetric form: $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z}{1}$. The denominators are the direction ratios. Since they are all $1$, the direction vector is indeed $(1, 1, 1)$. Always check the denominators carefully!

Q: I got $\lambda=0$ and $\mu=0$. Does that mean the points $Q$ and $R$ are just the points given in the line equations?

Exactly! When the parameter is zero, the general point $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$ simplifies to $(x_0, y_0, z_0)$. It is a beautiful simplification, but do not assume it will happen every time. Always solve the dot product equation to be sure.

Question

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 $PQR$ is $A$, then $4 A^2$ is equal to :

Options:

(A) $139$

(B) $147$

(C) $151$

(D) $143$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have a fixed point $P(5, 1, -3)$ and two distinct lines, $L_1$ and $L_2$, stretching out into the void.

Your mission is to drop perpendiculars from $P$ onto these lines, finding the exact points $Q$ and $R$ where these perpendiculars land. This is not just a calculation; it is a dance of vectors and parameters.

### The Parametric Bridge

First, we must tame these lines. A line in 3D is best understood through its parametric form.

For $L_1 : x - 1 = y - 2 = z$, we set this equal to a parameter $\lambda$. This allows us to express any point on $L_1$ as $Q(\lambda+1, \lambda+2, \lambda)$.

Similarly, for $L_2 : x - 2 = y = z - 1$, we introduce a new parameter $\mu$, giving us a general point $R(\mu+2, \mu, \mu+1)$. This is our bridge—we have turned a geometric line into an algebraic variable.

### The Perpendicularity Filter

Now, we find the exact location of $Q$ and $R$ using the power of the dot product. If the segment $\vec{PQ}$ is perpendicular to the line $L_1$, then the dot product of $\vec{PQ}$ and the direction vector of $L_1$ must be zero.

The direction vector of $L_1$ is $\vec{d_1} = (1, 1, 1)$. By calculating $\vec{PQ} = (\lambda-4, \lambda+1, \lambda+3)$ and setting $\vec{PQ} \cdot \vec{d_1} = 0$, we find:

$$3\lambda = 0 \Rightarrow \lambda = 0$$

This gives us $Q(1, 2, 0)$. We repeat this for $R$ on $L_2$ with $\vec{d_2} = (1, 1, 1)$, and again, we find $\mu = 0$, leading to $R(2, 0, 1)$.

### The Area Generator

With $P(5, 1, -3)$, $Q(1, 2, 0)$, and $R(2, 0, 1)$ in hand, we have formed a triangle in space. To find its area, we define two side vectors:

$$\vec{PQ} = -4\hat{i} + \hat{j} + 3\hat{k}$$
$$\vec{PR} = -3\hat{i} - \hat{j} + 4\hat{k}$$

The area $A$ of $	riangle PQR$ is given by $A = \frac{1}{2} |\vec{PQ} 	imes \vec{PR}|$. Calculating the cross product $\vec{PQ} 	imes \vec{PR}$ yields:

$$\vec{PQ} 	imes \vec{PR} = 7\hat{i} + 7\hat{j} + 7\hat{k}$$

The magnitude of this vector is $\sqrt{7^2 + 7^2 + 7^2} = 7\sqrt{3}$. Thus, the area is:

$$A = \frac{7\sqrt{3}}{2}$$

### The Victory Lap

Finally, we address the question's demand: $4A^2$. We take our area $A = \frac{7\sqrt{3}}{2}$, square it to get $\frac{49 	imes 3}{4}$, and multiply by $4$.

$$4A^2 = 4 	imes \left( \frac{49 	imes 3}{4} ight) = 49 	imes 3 = 147$$

The final result is **147**. Remember, in JEE Advanced, the complexity is often just a layer covering a beautiful, simple core.

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Visualized Solution

Visualizing the Setup

Foot of Perpendicular on $L_{1}$ (Setup)

Finding Coordinates of $Q$

Foot of Perpendicular on $L_{2}$ (Setup)

Finding Coordinates of $R$

Forming Triangle $P QR$

Defining Vectors $P Q$ and $P R$

Cross Product $P Q \times P R$

Magnitude of Cross Product

Calculating Area $A$

Final Evaluation of $4 A^{2}$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Bridge

The Perpendicularity Filter

The Area Generator

The Victory Lap

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Visualized Solution

Visualizing the Setup

Foot of Perpendicular on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ (Setup)

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

Foot of Perpendicular on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​ (Setup)

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

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

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

Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ​×PR