Let P be the foot of the perpendicular from the point Q(10,-3,-1) on the line (x-3)/7 = (y-2)/-1 = (

Q: Why do we introduce the parameter $t$ in the line equation?

A line in 3D space is not just a single point; it is an infinite collection of points. By setting the symmetric equation $\frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2}$ equal to a parameter $t$, we create a 'map' that allows us to represent any point on that line as a function of $t$. This turns a geometric problem into a simple algebraic one: we just need to find the specific value of $t$ that lands us on the point $P$.

Q: How do I know for sure that $\vec{PQ}$ is perpendicular to the line?

The problem defines $P$ as the 'foot of the perpendicular' from $Q$ to the line. By definition, the line segment connecting a point to a line at its foot is the shortest distance, and this segment is always orthogonal (perpendicular) to the line itself. Therefore, the vector $\vec{PQ}$ must be perpendicular to the line's direction vector $\vec{d}$.

Q: Can I use the cross product to find the area of the triangle?

Absolutely! The area of a triangle with vertices $P, Q, R$ is given by $\frac{1}{2} |\vec{PQ} \times \vec{PR}|$. However, in this specific problem, we discovered that $\vec{PQ} \cdot \vec{QR} = 0$, which means the triangle is right-angled at $Q$. Using $\frac{1}{2} \times \text{base} \times \text{height}$ is significantly faster and less prone to calculation errors than the cross product method.

Question

Let P be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$ Then the area of the right angled triangle PQR, where R is the point $(3,-2,1)$, is

Options:

(A) $9\sqrt{15}$

(B) $\sqrt{30}$

(C) $8\sqrt{15}$

(D) $3\sqrt{30}$

Sigma Percentile · Accepted Answer

### The Geometry of the Perpendicular

Welcome, fellow traveler in the world of 3D geometry! Today, we are going to unravel a classic JEE Advanced problem. It is not just about crunching numbers; it is about visualizing the elegant dance between points, lines, and vectors in three-dimensional space.

Imagine you are standing in a room. You have a straight wire (our line $L$) stretching across the room, and a point $Q$ floating in the air. You want to find the exact spot $P$ on that wire that is closest to $Q$. That is the 'foot of the perpendicular.'

Once we find $P$, we have a triangle $PQR$ to analyze. Let us begin.

### Phase 1: The Parametric Dance

To find the coordinates of $P$, we need a way to describe every point on the line $L$. We are given the symmetric form:

$$ \frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2} $$

By setting this equal to a parameter $t$, we unlock the secret to every point on this line. Any point $P$ on this line can be written as $P(3 + 7t, 2 - t, -1 - 2t)$.

This is our candidate for the foot of the perpendicular. As $t$ varies, $P$ slides along the line. We just need to find the specific $t$ that makes the segment $PQ$ perpendicular to the line.

### Phase 2: The Dot Product Revelation

Now, let us construct the vector $\vec{PQ}$. We subtract the coordinates of $P$ from $Q(10, -3, -1)$.

This gives us $\vec{PQ} = (10 - (3+7t), -3 - (2-t), -1 - (-1-2t))$, which simplifies to $\vec{PQ} = (7 - 7t, t - 5, 2t)$. We also know the direction vector of our line $L$ from the denominators of its equation: $\vec{d} = (7, -1, -2)$.

Here is the geometric soul of the problem: because $P$ is the foot of the perpendicular, $\vec{PQ}$ must be orthogonal to $\vec{d}$. This means their dot product must vanish: $\vec{PQ} \cdot \vec{d} = 0$.

Let us perform this calculation:

$$ 7(7 - 7t) - 1(t - 5) - 2(2t) = 0 $$

Expanding this, we get $49 - 49t - t + 5 - 4t = 0$, which simplifies to $54 - 54t = 0$. Solving this gives us the beautiful result $t = 1$.

### Phase 3: The Geometric Surprise

With $t = 1$, we can lock in the coordinates of $P$. Substituting $t = 1$ into our parametric form, we get $P(10, 1, -3)$.

Now, let us find the vectors that define our triangle. $\vec{PQ}$ becomes $(7 - 7(1), 1 - 5, 2(1)) = (0, -4, 2)$. The length of this height is:

$$ |\vec{PQ}| = \sqrt{0^2 + (-4)^2 + 2^2} = \sqrt{20} $$

Next, we find the vector $\vec{QR}$ using $R(3, -2, 1)$ and $Q(10, -3, -1)$. $\vec{QR} = R - Q = (-7, 1, 2)$. Its length is:

$$ |\vec{QR}| = \sqrt{(-7)^2 + 1^2 + 2^2} = \sqrt{54} $$

Before we calculate the area, let us check the angle at $Q$. The dot product $\vec{PQ} \cdot \vec{QR} = (0)(-7) + (-4)(1) + (2)(2) = 0 - 4 + 4 = 0$. The dot product is zero! Our triangle is right-angled at $Q$.

### Phase 4: The Final Calculation

Since we have a right-angled triangle, the area is simply:

$$ 	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes |\vec{QR}| 	imes |\vec{PQ}| $$

Substituting our values, we get:

$$ 	ext{Area} = \frac{1}{2} 	imes \sqrt{54} 	imes \sqrt{20} = \frac{1}{2} 	imes \sqrt{1080} $$

We can simplify $\sqrt{1080}$ as $\sqrt{36 	imes 30} = 6\sqrt{30}$. Finally:

$$ 	ext{Area} = \frac{1}{2} 	imes 6\sqrt{30} = 3\sqrt{30} $$

We have arrived at the solution. It is a testament to the power of vector algebra that such a complex 3D configuration collapses into such an elegant result. Keep practicing, and keep visualizing!

Animated Solution for Mathematics - Three Dimensional Geometry: Let P be the foot of the perpendicular from the point $Q (10, - 3, - 1)$ on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ Then the area of the right angled triangle PQR, where R is the point $(3, - 2, 1)$ , is

Select Answer:

Visualized Solution

The Geometric Setup

Parametric Form of Point $P$

Vector $P Q$

The Perpendicular Condition

Solving for Parameter $t$

Coordinates of Point $P$

Finding Vector $P Q$ and Length

Finding Vector $QR$ and Length

Verifying the Right Angle

Area of $△ P QR$

Final Result

The Sigma Insight: Equation of a Line in Space

The Geometry of the Perpendicular

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Consider the lines $L_{1} : x - 1 = y - 2 = z$ and $L_{2} : x - 2 = y = z - 1$ . Let the feet of the perpendiculars from the point $P (5, 1, - 3)$ on the lines $L_{1}$ and $L_{2}$ be $Q$ and $R$ respectively. If the area of the triangle $P QR$ is $A$ , then $4 A^{2}$ is equal to :

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

Let $P$ be the point $(10, - 2, - 1)$ and $Q$ be the foot of the perpendicular drawn from the point $R (1, 7, 6)$ on the line passing through the points $(2, - 5, 11)$ and $(- 6, 7, - 5)$ . Then the length of the line segment $P Q$ is equal to ________

Let the vertices $Q$ and $R$ of the triangle $P QR$ lie on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ , $QR = 5$ and the coordinates of the point $P$ be $(0, 2, 3)$ . If the area of the triangle $P QR$ is $\frac{m}{n}$ then :

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ , $λ \in R,$ intersect the line L at Q. Then $2 (P Q)^{2}$ is equal to:

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x + 2 y + z = 0$ and $3 y - z = 3$ is $(α, β, γ)$ , then $α + β + γ$ is equal to

The vertices B and C of a $Δ A B C$ lie on the line, $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A (1, - 1, 2)$ , is :-

Let P be the image of the point $Q (7, - 2, 5)$ in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R (5, p, q)$ be a point on L. Then the square of the area of $△ P QR$ is

Let a line pass through two distinct points $P (- 2, - 1, 3)$ and Q, and be parallel to the vector $3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ If the distance of the point Q from the point $R (1, 3, 3)$ is 5, then the square of the area of $△ P QR$ is equal to:

The vertices $B$ and $C$ of a triangle $A B C$ lie on the line $\frac{x}{1} = \frac{1 - y}{- 2} = \frac{z - 2}{3}$ . The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, α)$ respectively and $C$ is at a distance of 10 units from $B$ . The area (in sq. units) of $△ A B C$ is :