If two distinct point Q, R lie on the line of intersection of the planes -x + 2y - z = 0 and 3x - 5y

Q: Wait, why does the line of intersection pass through the origin?

Great observation! Look at the equations: $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$. If you plug in $(0,0,0)$ for $x, y,$ and $z$, both equations are satisfied ($0=0$). Since both planes contain the origin, their line of intersection must also pass through the origin. This simplifies our line equation significantly!

Q: Why is $T$ the midpoint of $QR$?

This is a classic property of isosceles triangles. In $\triangle PQR$, since $PQ = PR$, the triangle is isosceles with $P$ as the apex. The altitude from the apex to the base always bisects the base. Since $T$ is the foot of the perpendicular from $P$ to the line $L$ (which contains the base $QR$), $T$ must be the midpoint of $QR$.

Q: Can I just use the distance formula to find $Q$ and $R$?

You could, but it would be much more tedious! You would have to set the distance $PQ = \sqrt{18}$ and solve a quadratic equation for $\lambda$. While that works, finding the foot of the perpendicular $T$ first is a much more elegant and standard approach in 3D geometry. It breaks the problem into smaller, manageable steps: finding the height $PT$ and then the base $QR$.

Question

If two distinct point $Q, R$ lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$ and $PQ = PR = \sqrt{18}$ where the point $P$ is $(1, -2, 3)$, then the area of the triangle $PQR$ is equal to

Options:

(A) $\frac{2}{3}\sqrt{38}$

(B) $\frac{4}{3}\sqrt{38}$

(C) $\frac{8}{3}\sqrt{38}$

(D) $\sqrt{\frac{152}{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You are presented with two planes, $P_1: -x + 2y - z = 0$ and $P_2: 3x - 5y + 2z = 0$. These planes are like two infinite sheets of paper slicing through space.

When they meet, they collide to form a perfect, infinite straight line, which we will call $L$. Since the line $L$ lies on both planes, it must be perpendicular to the normal vectors of both planes.

The normal vector of $P_1$ is $\vec{n}_1 = -\hat{i} + 2\hat{j} - \hat{k}$, and the normal vector of $P_2$ is $\vec{n}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k}$. To find a vector $\vec{d}$ perpendicular to both, we use the cross product:

$$\vec{d} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & -1 \ 3 & -5 & 2 \end{vmatrix}$$

Calculating this determinant, we get $\vec{d} = \hat{i}(4 - 5) - \hat{j}(-2 + 3) + \hat{k}(5 - 6) = -\hat{i} - \hat{j} - \hat{k}$. For simplicity, we can scale this to $\vec{d} = \hat{i} + \hat{j} + \hat{k}$.

Because the planes have no constant terms, they pass through the origin $(0,0,0)$, and thus our line $L$ also passes through the origin. The equation of our line is simply:

$$\frac{x}{1} = \frac{y}{1} = \frac{z}{1} = \lambda$$

### The Isosceles Insight

Now, we introduce point $P(1, -2, 3)$. We are told that points $Q$ and $R$ lie on $L$ such that $PQ = PR = \sqrt{18}$. This creates an isosceles triangle $	riangle PQR$ with the base $QR$ resting on line $L$.

The beauty of an isosceles triangle is that the altitude from the apex $P$ to the base $QR$ bisects the base. Let $T$ be the foot of this altitude. $T$ is the point on $L$ closest to $P$.

Since $T$ lies on $L$, its coordinates are $(\lambda, \lambda, \lambda)$. The vector $\vec{PT}$ is given by:

$$\vec{PT} = T - P = (\lambda - 1, \lambda + 2, \lambda - 3)$$

Because $PT$ is the altitude, $\vec{PT}$ must be perpendicular to the line's direction $\vec{d} = (1, 1, 1)$. Thus, their dot product must be zero:

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

Solving this, we get $3\lambda - 2 = 0$, so $\lambda = \frac{2}{3}$. The foot of the perpendicular is $T(\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$.

### The Final Calculation

Now, we calculate the height $PT$. The squared distance $PT^2$ is:

$$PT^2 = \left(\frac{2}{3} - 1ight)^2 + \left(\frac{2}{3} + 2ight)^2 + \left(\frac{2}{3} - 3ight)^2 = \left(-\frac{1}{3}ight)^2 + \left(\frac{8}{3}ight)^2 + \left(-\frac{7}{3}ight)^2$$

$$PT^2 = \frac{1 + 64 + 49}{9} = \frac{114}{9} = \frac{38}{3}$$

In the right-angled triangle $	riangle PQT$, we have $PQ^2 = PT^2 + QT^2$. Substituting $PQ^2 = 18$ and $PT^2 = \frac{38}{3}$, we find:

$$QT^2 = 18 - \frac{38}{3} = \frac{54 - 38}{3} = \frac{16}{3} \Rightarrow QT = \frac{4}{\sqrt{3}}$$

Since $T$ is the midpoint of the base, the total base $QR = 2 	imes QT = \frac{8}{\sqrt{3}}$. Finally, the area of $	riangle PQR$ is:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{8}{\sqrt{3}} 	imes \sqrt{\frac{38}{3}} = \frac{4}{\sqrt{3}} 	imes \frac{\sqrt{38}}{\sqrt{3}} = \mathbf{\frac{4}{3}\sqrt{38}}$$

Animated Solution for Mathematics - Three Dimensional Geometry: If two distinct point $Q, R$ lie on the line of intersection of the planes $- x + 2 y - z = 0$ and $3 x - 5 y + 2 z = 0$ and $P Q = P R = 18$ where the point $P$ is $(1, - 2, 3)$ , then the area of the triangle $P QR$ is equal to

Select Answer:

Visualized Solution

Visualizing the Intersection

Direction of the Intersection Line

Calculating the Direction Vector

Equation of Line $L$

The Isosceles Triangle

Foot of the Perpendicular

Setting up the Dot Product

Coordinates of $T$

Calculating Height $P T$

Finding the Base $QR$

Calculating $QT$

Area of Triangle $P QR$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Isosceles Insight

The Final Calculation

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