Let A and B be two distinct points on the line L: (x-6)/3 = (y-7)/2 = (z-7)/-2. Both A and B are at

Q: Why can't we just use the distance formula from $P$ to a general point on the line to find $A$ and $B$?

You certainly could, but that would lead to a quadratic equation in $\lambda$ that is much more tedious to solve. By finding the foot of the perpendicular $Q$ first, you decompose the problem into two simpler, manageable steps: finding the closest point and then moving along the line.

Q: How do I know if I should add or subtract the direction vector?

The problem asks for two distinct points $A$ and $B$ at a distance $2\sqrt{17}$ from $Q$. Since the line extends in two directions, the $\pm$ sign naturally accounts for both points. One point is 'ahead' of $Q$, and the other is 'behind' it.

Q: Is there a shortcut for $\vec{OA} \cdot \vec{OB}$?

While there are properties of dot products, calculating the coordinates of $A$ and $B$ is the most robust method here. Given the numbers simplify so cleanly, direct calculation is actually the fastest and least error-prone path to the answer.

Question

Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$. If $O$ is the origin, then $\vec{OA} \cdot \vec{OB}$ is equal to:

Options:

(A) $49$

(B) $47$

(C) $21$

(D) $62$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast room. There is a straight line $L$ stretching across the room, and a point $P(1, 2, 3)$ hovering in space. We are tasked with finding two points, $A$ and $B$, on this line that are defined by their relationship to the 'foot of the perpendicular' from $P$ to $L$.

Let's call this foot $Q$. Think of $Q$ as the shadow of $P$ cast onto the line $L$. To find $Q$, we use the parametric form of the line. By setting the line equation equal to a parameter $\lambda$:

$$\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda$$

We can express any point on the line as $Q(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$. This is our anchor.

### The Perpendicularity Condition

Now, we must pin down $\lambda$. We know that the vector $\vec{PQ}$ must be perpendicular to the line $L$. In the language of vectors, this means the dot product of $\vec{PQ}$ and the direction vector of the line, $\vec{d} = (3, 2, -2)$, must be zero.

That is, $\vec{PQ} \cdot \vec{d} = 0$. First, we calculate the vector $\vec{PQ}$:

$$\vec{PQ} = (3\lambda + 6 - 1, 2\lambda + 7 - 2, -2\lambda + 7 - 3) = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4)$$

Taking the dot product with $\vec{d} = (3, 2, -2)$, we get the following equation:

$$3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0$$

Expanding this, we find:

$$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$$

This simplifies to $17\lambda + 17 = 0$, which yields $\lambda = -1$. Substituting this back into our expression for $Q$, we find our anchor point $Q(3, 5, 9)$.

### The Leap Along the Line

With $Q$ found, the rest of the journey is a simple walk. We need points $A$ and $B$ at a distance of $2\sqrt{17}$ from $Q$. The direction vector $\vec{d} = (3, 2, -2)$ has a magnitude of:

$$|\vec{d}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}$$

This is a gift! It means that moving one unit of $\vec{d}$ takes us $\sqrt{17}$ units of distance. To move $2\sqrt{17}$ units, we simply move $2\vec{d}$ from $Q$.

Thus, $A$ and $B$ are given by $Q \pm 2\vec{d}$:

$$A = (3 + 2(3), 5 + 2(2), 9 + 2(-2)) = (9, 9, 5)$$

$$B = (3 - 2(3), 5 - 2(2), 9 - 2(-2)) = (-3, 1, 13)$$

### The Final Calculation

We have arrived at the finish line. We need the dot product $\vec{OA} \cdot \vec{OB}$. Since $O$ is the origin, $\vec{OA}$ and $\vec{OB}$ are simply the coordinates of $A$ and $B$.

We compute the dot product as follows:

$$\vec{OA} \cdot \vec{OB} = (9)(-3) + (9)(1) + (5)(13)$$

$$\vec{OA} \cdot \vec{OB} = -27 + 9 + 65 = 47$$

The final result is **47**. It is a beautiful, clean integer—the hallmark of a well-crafted JEE problem.

Select Answer:

Visualized Solution

Visualizing the Geometry

Defining General Point $Q$

Vector $P Q$ and Direction $d$

Applying Perpendicularity

Solving for $λ$

Coordinates of Foot $Q$

Finding Points $A$ and $B$

Calculating Coordinates of $A$ and $B$

The Final Dot Product

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Perpendicularity Condition

The Leap Along the Line

The Final Calculation

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