Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If

Q: Wait, why is $\triangle PQR$ right-angled at $Q$?

Great question! By definition, $Q$ is the foot of the perpendicular from $P$ to the plane. This means the line segment $PQ$ is perpendicular to the plane itself. Since the line $QR$ lies entirely within the plane, $PQ$ must be perpendicular to $QR$. Thus, $\angle PQR = 90^\circ$.

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

You absolutely could, but it is not necessary! Finding the coordinates of $Q$ involves solving for the line passing through $P$ and perpendicular to the plane, then finding its intersection with the plane. That is a valid but longer path. Using the perpendicular distance formula directly is much more elegant and efficient.

Q: Why is the area $\frac{1}{2} \times PQ \times QR$?

In any right-angled triangle, the two sides forming the right angle can be considered the base and the height. Since $\angle PQR = 90^\circ$, $PQ$ and $QR$ are the legs of the triangle. Therefore, the area is simply $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times QR \times PQ$.

Question

Let $Q$ be the foot of perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^\circ$, then the area of $	riangle PQR$ is equal to:

Options:

(A) $\frac{\sqrt{3}}{2}$

(B) $\sqrt{3}$

(C) $2\sqrt{3}$

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. Before you lies an infinite, flat surface—a plane defined by the elegant equation $x + 2y + z = 14$.

Floating above this plane, like a star in the night sky, is a point $P$ with coordinates $(1, 2, 3)$. This is the setup for our problem, and it is a beautiful example of how 3D geometry allows us to bridge the gap between abstract equations and physical reality.

Our goal is to find the area of a triangle formed by this point $P$, the foot of the perpendicular $Q$ dropped from $P$ onto the plane, and an arbitrary point $R$ on the plane that satisfies a specific condition: $\angle PRQ = 60^{\circ}$.

### The Perpendicular: Finding the Altitude

The first step in our journey is to find the length of the segment $PQ$. This segment is the shortest distance from the point $P$ to the plane, which serves as the altitude of our triangle $	riangle PQR$.

We use the classic perpendicular distance formula:

$$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Substituting our values, where the plane is $x + 2y + z - 14 = 0$ and the point is $(1, 2, 3)$, we get:

$$PQ = \frac{|1(1) + 2(2) + 1(3) - 14|}{\sqrt{1^2 + 2^2 + 1^2}}$$

Calculating the numerator, we have $1 + 4 + 3 - 14 = 8 - 14 = -6$. Taking the absolute value, we get $6$. The denominator is $\sqrt{1 + 4 + 1} = \sqrt{6}$.

Thus, $PQ = \frac{6}{\sqrt{6}} = \sqrt{6}$. We have successfully found the height of our triangle!

### The Trigonometric Bridge: Unlocking the Base

Now that we have the height $PQ = \sqrt{6}$, we turn our attention to the triangle $	riangle PQR$. Because $PQ$ is perpendicular to the plane, it is perpendicular to any line in the plane that passes through $Q$.

Therefore, $	riangle PQR$ is a right-angled triangle with the right angle at $Q$. We are given that $\angle PRQ = 60^{\circ}$.

In this right-angled triangle, we know the opposite side $PQ$ and the angle at $R$. We need the base $QR$. The tangent function is our best friend here:

$$	an(60^{\circ}) = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{PQ}{QR}$$

We know $	an(60^{\circ}) = \sqrt{3}$. So, $\sqrt{3} = \frac{\sqrt{6}}{QR}$. Solving for $QR$, we find:

$$QR = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$$

### The Final Calculation: The Area of $	riangle PQR$

We have reached the final phase of our journey. We have the height $PQ = \sqrt{6}$ and the base $QR = \sqrt{2}$.

The area of a right-angled triangle is given by the formula:

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

Substituting our values, we get:

$$	ext{Area} = \frac{1}{2} 	imes \sqrt{2} 	imes \sqrt{6}$$

Multiplying the square roots, we get $\sqrt{12}$, which simplifies to $2\sqrt{3}$. Thus, the final area is:

$$	ext{Area} = \frac{1}{2} 	imes 2\sqrt{3} = \mathbf{\sqrt{3}}$$

The elegance of this result is truly satisfying. We started with a point and a plane in 3D space, and through the power of geometry and trigonometry, we arrived at a clean, precise area.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $Q$ be the foot of perpendicular drawn from the point $P (1, 2, 3)$ to the plane $x + 2 y + z = 14$ . If $R$ is a point on the plane such that $∠ P R Q = 6 0^{\circ}$ , then the area of $△ P QR$ is equal to:

Select Answer:

Visualized Solution

Visualize the Setup

Drop the Perpendicular

Introduce Point $R$

The Distance Formula

Raw Setup for $P Q$

Compute Numerator and Denominator

Finalize $P Q$

Trigonometry in $△ P QR$

Raw Setup for $QR$

Substitute and Compute $QR$

Area Formula

Raw Setup for Area

Final Compute

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Perpendicular

The Trigonometric Bridge

The Final Calculation

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