Let the line L: (x-1)/2 = (y+1)/-1 = (z-3)/1 intersect the plane 2x + y + 3z = 16 at the point P. Le

Q: Why do we use two different parameters, $\lambda$ and $\mu$?

Great question! We use $\lambda$ to find the intersection point $P$ and $\mu$ to find the foot of the perpendicular $Q$. Since $P$ and $Q$ are generally different points on the line, they correspond to different values of the parameter. Using the same parameter would incorrectly force $P$ and $Q$ to be the same point.

Q: Why is the dot product of $\vec{RQ}$ and the direction vector zero?

Because $\vec{RQ}$ is the vector representing the perpendicular line segment from $R$ to the line $L$. By definition, if a vector is perpendicular to a line, it must be perpendicular to the line's direction vector. The dot product of two perpendicular vectors is always zero.

Q: Is there a faster way to find the area of the triangle?

While you could use the base-height formula, the cross product method $\frac{1}{2} |\vec{QP} \times \vec{QR}|$ is the most robust and standard approach in 3D geometry. It directly uses the coordinates of the vertices, minimizing the chance of errors in calculating the height.

Question

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x + y + 3z = 16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1, -1, -3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$, then $\alpha^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are exploring the architecture of three-dimensional space.

Imagine standing in a room where a laser beam, our line $L$, cuts through a sheet of paper, our plane. The point where they meet, $P$, is the first piece of our puzzle.

To find it, we embrace the power of parameterization. By setting the line's equation equal to $\lambda$, we transform a complex 3D coordinate into a simple, manageable expression: $(2\lambda+1, -\lambda-1, \lambda+3)$.

This is the key that unlocks the door. When we substitute this into the plane's equation:

$$2(2\lambda+1) + (-\lambda-1) + 3(\lambda+3) = 16$$

We are essentially asking the line, "Where do you touch the plane?" The algebra unfolds as follows:

$$6\lambda + 10 = 16$$

This leads us to $\lambda = 1$, and thus, the coordinates of point $P$ are:

$$P = (3, -2, 4)$$

### The Foot of the Perpendicular

Now, we shift our focus to the second act: the foot of the perpendicular. We have a point $R(1, -1, -3)$ and we need to drop a perpendicular to the line $L$.

We define a general point $Q$ on the line using a new parameter $\mu$. The vector $\vec{RQ}$ connects our point $R$ to this general point $Q$.

The magic happens when we invoke the condition of perpendicularity: the dot product of $\vec{RQ}$ and the direction vector of the line $\vec{d} = (2, -1, 1)$ must be zero. This condition, $\vec{RQ} \cdot \vec{d} = 0$, is the heartbeat of this problem.

Solving the resulting linear equation:

$$4\mu + \mu + \mu + 6 = 0$$

This yields $\mu = -1$, revealing the coordinates of point $Q$:

$$Q = (-1, 0, 2)$$

### The Final Area Calculation

Finally, we stand before the triangle $PQR$. With vertices $P(3, -2, 4)$, $Q(-1, 0, 2)$, and $R(1, -1, -3)$, we find the vectors $\vec{QP}$ and $\vec{QR}$.

The area $\alpha$ is half the magnitude of their cross product. The calculation is defined as:

$$\alpha = \frac{1}{2} |\vec{QP} 	imes \vec{QR}|$$

Performing the cross product and calculating the magnitude leads us to the final result:

$$\alpha^2 = 180$$

You have navigated the geometry, mastered the algebra, and conquered the 3D space. **The final value is 180.**

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Parametric Form of Line $L$

Finding Intersection Point $P$

Solving for $λ$

Coordinates of Point $P$

Locating Point $Q$

Vector $R Q$ and Direction of $L$

Dot Product Condition

Solving for $μ$

Coordinates of Point $Q$

Vectors $QP$ and $QR$

Cross Product for Area

Final Calculation of $α^{2}$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Foot of the Perpendicular

The Final Area Calculation

Similar Questions

Let $d$ be the distance between the foot of perpendiculars of the points $P (1, 2, - 1)$ and $Q (2, - 1, 3)$ on the plane $- x + y + z = 1$ . Then $d^{2}$ is equal to ____.

If the line $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z - 1}{- 1}$ intersects the plane $2 x + 3 y - z + 13 = 0$ at a point P and the plane $3 x + y + 4 z = 16$ at a point Q, then PQ is equal to :

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 $5 x - 4 y - z = 1$ . 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 $P S$ is

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersects at the point $P$ , then the distance of the point $P$ from the plane $z = a$ is :

If the line, $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}$ lies in the plane, $l x + m y - z = 9$ , then $l^{2} + m^{2}$ is equal to :

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

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

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

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

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

Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RQ​ and Direction of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L