If Q(0, -1, -3) is the image of the point P in the plane 3x - y + 4z = 2 and R is the point (3, -1,

Q: Why is the line $PQ$ perpendicular to the plane? Isn't that just an assumption?

It is the definition of a reflection! In 3D geometry, the image of a point in a plane is defined such that the line connecting the point and its image is perpendicular to the plane, and the plane bisects that segment. It is the fundamental geometric property of mirrors.

Q: Do I really need to find the coordinates of $P$ to solve this?

Absolutely not! That is a common trap. Finding $P$ involves solving linear equations and is prone to calculation errors. By using the vector properties of the normal and the cross product, we bypass the need to know $P$ entirely. Always look for the vector shortcut.

Q: Why does the $\sqrt{26}$ cancel out so perfectly in the final step?

That is the beauty of JEE Advanced problems! They are designed with internal symmetry. When you see terms like $\sqrt{26}$ appearing in both the base and the height, it is a sign that you are on the right track. It is the problem's way of rewarding your correct geometric intuition.

Question

If $Q(0, -1, -3)$ is the image of the point $P$ in the plane $3x - y + 4z = 2$ and $R$ is the point $(3, -1, -2)$, then the area (in sq. units) of $\Delta PQR$ is :

Options:

(A) $\sqrt{65}/2$

(B) $\sqrt{91}/4$

(C) $2\sqrt{13}$

(D) $\sqrt{91}/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a mirror. You see your reflection, and you know instinctively that the line connecting you to your reflection is perpendicular to the mirror's surface. This is the exact physical intuition we need to solve this problem.

We are given a plane $3x - y + 4z = 2$ and a point $Q(0, -1, -3)$, which is the image of a point $P$. The plane acts as our mirror.

### The Geometry of Reflection

Because $Q$ is the image of $P$, the line segment $PQ$ must be perpendicular to the plane. This is our first breakthrough.

The direction of this line $PQ$ is identical to the normal vector of the plane, which we can extract directly from the plane's equation:
$$\vec{n} = 3\hat{i} - \hat{j} + 4\hat{k}$$

We do not need to find the coordinates of $P$ to know the direction of the line $PQ$. This saves us precious time and reduces the chance of algebraic errors.

### The Base of the Triangle

To find the area of $\Delta PQR$, we need the base $PQ$. We know that the plane bisects the segment $PQ$. Therefore, the length of $PQ$ is exactly twice the perpendicular distance from $Q$ to the plane.

Let's calculate this distance $d$ using the standard formula:
$$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Substituting our point $Q(0, -1, -3)$ and the plane $3x - y + 4z - 2 = 0$, we get:
$$d = \frac{|3(0) - (-1) + 4(-3) - 2|}{\sqrt{3^2 + (-1)^2 + 4^2}}$$

Simplifying the numerator, we get $|0 + 1 - 12 - 2| = |-13| = 13$. The denominator is $\sqrt{9 + 1 + 16} = \sqrt{26}$. Thus:
$$d = \frac{13}{\sqrt{26}} = \frac{\sqrt{26}}{2}$$

Since $PQ = 2d$, we find that the base $PQ = \sqrt{26}$.

### The Height Challenge

Now we introduce point $R(3, -1, -2)$. We need the perpendicular height $h$ from $R$ to the line $PQ$. This is where vector algebra shines.

The height $h$ is the perpendicular distance from a point to a line. The formula is:
$$h = \frac{|\vec{QR} 	imes \vec{n}|}{|\vec{n}|}$$

First, let's find vector $\vec{QR} = (3-0)\hat{i} + (-1 - (-1))\hat{j} + (-2 - (-3))\hat{k} = 3\hat{i} + 0\hat{j} + \hat{k}$. Now, we compute the cross product $\vec{V} = \vec{QR} 	imes \vec{n}$:
$$\vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 0 & 1 \ 3 & -1 & 4 \end{vmatrix}$$

Expanding this determinant, we get $\hat{i}(0 - (-1)) - \hat{j}(12 - 3) + \hat{k}(-3 - 0) = \hat{i} - 9\hat{j} - 3\hat{k}$. The magnitude of this vector is:
$$|\vec{V}| = \sqrt{1^2 + (-9)^2 + (-3)^2} = \sqrt{1 + 81 + 9} = \sqrt{91}$$

### The Grand Finale

We are almost there! The height $h$ is $\frac{\sqrt{91}}{\sqrt{26}}$. The area of $\Delta PQR$ is:
$$	ext{Area} = \frac{1}{2} 	imes 	ext{Base} 	imes 	ext{Height} = \frac{1}{2} 	imes \sqrt{26} 	imes \frac{\sqrt{91}}{\sqrt{26}}$$

Notice the elegance here? The $\sqrt{26}$ terms cancel out perfectly, leaving us with the final result:
$$	ext{Area} = \frac{\sqrt{91}}{2}$$

This is the beauty of mathematics—when you follow the logical path, the complexity often dissolves into a simple, elegant result. You have successfully navigated the 3D geometry of reflections and vectors.

Animated Solution for Mathematics - Three Dimensional Geometry: If $Q (0, - 1, - 3)$ is the image of the point $P$ in the plane $3 x - y + 4 z = 2$ and $R$ is the point $(3, - 1, - 2)$ , then the area (in sq. units) of $Δ P QR$ is :

Select Answer:

Visualized Solution

The Geometric Setup

Understanding the Image Relationship

Distance of $Q$ from the Plane

Calculating the Distance

Simplifying the Distance

Finding the Base $P Q$

Introducing Point $R$

Direction of Line $P Q$

Vector $QR$ Calculation

Cross Product for Height

Expanding the Determinant

Magnitude of the Cross Product

Calculating the Height $h$

The Area Formula

Final Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Geometry of Reflection

The Base of the Triangle

The Height Challenge

The Grand Finale

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