The point P is the intersection of the straight line joining the points Q(2, 3, 5) and R(1, -1, 4) w

Q: Wait, why do we use $\lambda$ for $P$ and $\mu$ for $S$? Can't we just use one variable?

That is a fantastic question! Think of it this way: $P$ and $S$ are two completely different locations on the line. If you force them to share the same parameter, you are essentially saying $P = S$, which is almost certainly false. By using $\lambda$ and $\mu$, you give each point the freedom to find its own unique 'address' on the line.

Q: Why does the dot product of $\vec{TS}$ and the line's direction vector have to be zero?

Because that is the definition of perpendicularity in vector algebra! If two vectors are perpendicular, the angle between them is $90^\circ$. Since $\vec{A} \cdot \vec{B} = |A||B| \cos(\theta)$, and $\cos(90^\circ) = 0$, the dot product must vanish. It is the most elegant way to enforce the 'shortest distance' condition.

Q: Is there a shortcut? This feels like a lot of algebra.

In JEE Advanced, accuracy is your best shortcut. While there are formulas for the foot of the perpendicular, deriving them on the fly is risky. The method of parameterization is robust, foolproof, and less prone to memory errors. Trust the process!

Question

The point $P$ is the intersection of the straight line joining the points $Q(2, 3, 5)$ and $R(1, -1, 4)$ with the plane $5x - 4y - z = 1$. If $S$ is the foot of the perpendicular drawn from the point $T(2, 1, 4)$ to $QR$, then the length of the line segment $PS$ is

Options:

(A) $\frac{1}{\sqrt{2}}$

(B) $\sqrt{2}$

(C) 2

(D) $2\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Architecture of 3D Space: A Journey to $P$ and $S$

Welcome, future engineer. Today, we are not just solving a problem; we are navigating the architecture of 3D space. Often, students look at 3D geometry and see a mess of coordinates. I want you to see it as a landscape.

We have a line, a plane, and a point. Our mission is to find two specific landmarks: $P$ and $S$. Let us begin by defining the path.

### Phase 1: The Parameterized Path

The line $QR$ is our highway. To travel along it, we need a direction. By subtracting the coordinates of $Q(2, 3, 5)$ from $R(1, -1, 4)$, we find the direction vector $\vec{d} = (-1, -4, -1)$.

Now, we introduce the parameter $\lambda$. This is the most powerful tool in your 3D toolkit. By writing the line as:

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

We effectively say: "Give me any value of $\lambda$, and I will give you a point on the line." Our general point is $(2 - \lambda, 3 - 4\lambda, 5 - \lambda)$. This is the key that unlocks the entire problem.

### Phase 2: The Intersection at $P$

Now, imagine the plane $5x - 4y - z = 1$ slicing through our line. Point $P$ is the exact moment of impact.

Since $P$ lies on the line, it must have coordinates $(2 - \lambda, 3 - 4\lambda, 5 - \lambda)$. Since it also lies on the plane, it must satisfy the plane's equation. We substitute our general point into the plane equation:

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

As we expand this, watch the terms dance: $10 - 5\lambda - 12 + 16\lambda - 5 + \lambda = 1$. Simplifying this, we get $12\lambda - 7 = 1$, which leads us to $\lambda = \frac{2}{3}$.

Substituting this back, we find $P = (\frac{4}{3}, \frac{1}{3}, \frac{13}{3})$. We have our first landmark.

### Phase 3: The Foot of the Perpendicular $S$

Now, shift your focus to point $T(2, 1, 4)$. We are dropping a perpendicular from $T$ to the line $QR$. The point where it lands is $S$.

Because $S$ is also on the line, it must have its own parameter, $\mu$. Thus, $S = (2 - \mu, 3 - 4\mu, 5 - \mu)$. We need the vector $\vec{TS}$, which is $S - T = (-\mu, 2 - 4\mu, 1 - \mu)$.

Here is the beauty of vector geometry: for $TS$ to be the foot of the perpendicular, $\vec{TS}$ must be orthogonal to the line $QR$. This means their dot product must be zero: $\vec{TS} \cdot \vec{d}_{QR} = 0$.

Calculating $(-\mu)(-1) + (2 - 4\mu)(-4) + (1 - \mu)(-1) = 0$, we get $\mu - 8 + 16\mu - 1 + \mu = 0$, which simplifies to $18\mu = 9$, or $\mu = \frac{1}{2}$. Substituting $\mu = \frac{1}{2}$ into our general point, we find $S = (\frac{3}{2}, 1, \frac{9}{2})$.

### Phase 4: The Final Distance

The journey is almost over. We have $P(\frac{4}{3}, \frac{1}{3}, \frac{13}{3})$ and $S(\frac{3}{2}, 1, \frac{9}{2})$. The distance formula is our final bridge:

$$PS = \sqrt{\left(\frac{4}{3} - \frac{3}{2}\right)^2 + \left(\frac{1}{3} - 1\right)^2 + \left(\frac{13}{3} - \frac{9}{2}\right)^2}$$

Calculating the differences, we get $(-\frac{1}{6})^2 + (-\frac{2}{3})^2 + (-\frac{1}{6})^2$. This becomes:

$$\frac{1}{36} + \frac{4}{9} + \frac{1}{36} = \frac{1 + 16 + 1}{36} = \frac{18}{36} = \frac{1}{2}$$

Taking the square root, we arrive at **$\frac{1}{\sqrt{2}}$**. You have successfully navigated the 3D landscape. Remember, in JEE Advanced, it is not just about the calculation; it is about the visualization. You did well.

Select Answer:

Visualized Solution

Visualizing the Setup

Direction Ratios of $QR$

Equation of Line $QR$

General Point on Line $QR$

Intersection with the Plane

Solving for $λ$

Coordinates of Point $P$

Foot of the Perpendicular $S$

Vector $T S$

The Perpendicular Condition

Solving for $μ$

Coordinates of Point $S$

Distance Formula for $P S$

Calculating the Distance

Final Result

The Sigma Insight: Intersection of a Line and a Plane

The Architecture of 3D Space

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Visualizing the Setup

Direction Ratios of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

General Point on Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Intersection with the Plane

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

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

Foot of the Perpendicular .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

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