Let the vertices Q and R of the triangle PQR lie on the line (x+3)/5 = (y-1)/2 = (z+4)/3, QR = 5 and

Q: Why do we use the parameter $\lambda$ instead of just using the distance formula from a point to a line?

While there is a standard formula for the distance from a point to a line, deriving it using the parameter $\lambda$ and the dot product condition is a fundamental technique in JEE Advanced. It builds your geometric intuition and ensures you don't have to memorize complex formulas, which can be prone to errors under exam pressure.

Q: Does the position of $Q$ and $R$ on the line affect the area?

Great question! No, it does not. The problem states that $QR = 5$. Since the line is infinite, $Q$ and $R$ can slide anywhere along the line, but as long as the distance between them is fixed at $5$, the base of the triangle remains constant. The height from $P$ to the line is also constant. Therefore, the area remains invariant regardless of where $Q$ and $R$ are placed.

Q: What if the line was given in a different form, like the intersection of two planes?

If the line is given as the intersection of two planes, your first step should always be to convert it into the symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. You can do this by finding the direction vector $\vec{v} = \vec{n_1} \times \vec{n_2}$ and finding any point on the line by setting one variable to zero. Once you have the symmetric form, the method used here remains exactly the same.

Question

Let the vertices $Q$ and $R$ of the triangle $PQR$ 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 $PQR$ is $\frac{m}{n}$ then :

Options:

(A) $m - 5\sqrt{21}n = 0$

(B) $2m - 5\sqrt{21}n = 0$

(C) $5m - 2\sqrt{21}n = 0$

(D) $5m - 21\sqrt{2}n = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of 3D geometry.

Imagine standing in a vast, empty room. You have a straight, infinite wire stretched across the room, and a single point $P$ floating in the air. We are tasked with forming a triangle $PQR$ where $Q$ and $R$ are on the wire, and $P$ is our apex.

The base $QR$ is fixed at $5$ units. Our goal is to find the area of this triangle.

### The Strategy: Breaking Down the Complexity

The beauty of this problem lies in its simplicity. We know the area of a triangle is given by:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

We already have the base, $QR = 5$. The only missing piece of our puzzle is the height.

In 3D geometry, the height of a triangle from a point to a line is the perpendicular distance from that point to the line. Let's call the foot of this perpendicular $M$. Our mission is to find the length of the segment $PM$.

### The Parametric Leap: Unlocking the Line

How do we find $M$? We know $M$ lies on the line $L$ defined by:

$$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$

This is where the magic of the parameter $\lambda$ comes in. By setting this equation equal to $\lambda$, we can express any point on the line as a function of $\lambda$.

Thus, the coordinates of $M$ are:

$$M = (5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$$

This is our key to the kingdom. Every point on that line is now captured by a single variable.

### The Orthogonality Condition: The Dot Product

Now, consider the vector $\vec{PM}$. It connects our apex $P(0, 2, 3)$ to the foot of the perpendicular $M$.

Since $PM$ is perpendicular to the line, the vector $\vec{PM}$ must be orthogonal to the line's direction vector $\vec{v} = (5, 2, 3)$. The condition for orthogonality is simple yet profound: the dot product must be zero.

$$\vec{PM} \cdot \vec{v} = 0$$

Let's calculate $\vec{PM} = M - P$:

$$\vec{PM} = (5\lambda - 3 - 0, 2\lambda + 1 - 2, 3\lambda - 4 - 3) = (5\lambda - 3, 2\lambda - 1, 3\lambda - 7)$$

Now, we apply the dot product:

$$5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0$$

### The Algebraic Triumph

Let's expand this carefully:

$$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$$

Combining the terms, we find:

$$38\lambda - 38 = 0 \Rightarrow \lambda = 1$$

With $\lambda = 1$, we can find the exact coordinates of $M$:

$$M = (5(1) - 3, 2(1) + 1, 3(1) - 4) = (2, 3, -1)$$

Now, the distance $PM$ is the magnitude of the vector $\vec{PM} = (2-0, 3-2, -1-3) = (2, 1, -4)$. The length $h$ is:

$$h = |\vec{PM}| = \sqrt{2^2 + 1^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$$

### The Final Celebration

We have the base $5$ and the height $\sqrt{21}$. The area is:

$$	ext{Area} = \frac{1}{2} 	imes 5 	imes \sqrt{21} = \frac{5\sqrt{21}}{2}$$

The problem defines this as $\frac{m}{n}$, so $\frac{m}{n} = \frac{5\sqrt{21}}{2}$.

We have conquered the problem! Remember, in JEE Advanced, it's not just about the answer; it's about the clarity of your thought process. Keep practicing, keep visualizing, and keep falling in love with the mathematics behind the scenes.

Select Answer:

Visualized Solution

Visualizing the Setup

Triangle $P QR$

The Height of the Triangle

Parametric Coordinates of $M$

Defining Vector $P M$

Direction Vector of Line $L$

The Perpendicularity Condition

Applying the Dot Product

Solving for $λ$

Coordinates of $M$

Calculating the Height $h$

Calculating the Area

Final Relation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Strategy

The Parametric Leap

The Orthogonality Condition

The Algebraic Triumph

The Final Celebration

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