Let P be the point (10, -2, -1) and Q be the foot of the perpendicular drawn from the point R(1, 7,

Q: Why do we set the dot product to zero? Can't we just use the distance formula and minimize it?

You absolutely can! Minimizing the distance squared is a valid approach, but it involves calculus (differentiation). The dot product method is the 'geometric shortcut'—it's faster, cleaner, and less prone to algebraic errors. It directly uses the property that the shortest distance from a point to a line is along the perpendicular, where the angle between the line and the segment is $90^\circ$, making the dot product zero.

Q: Does the direction vector $\vec{d}$ have to be $(2, -3, 4)$? Could I use $(-2, 3, -4)$?

Great observation! Yes, you can use any scalar multiple of the direction vector. The dot product $\vec{RQ} \cdot \vec{d} = 0$ will yield the same $\lambda$ regardless of whether you use $\vec{d}$ or $-\vec{d}$. The line is defined by its direction, not its orientation, so any vector parallel to the line works perfectly.

Q: I got $\lambda = -2$, but what if I made a sign error? How do I check my answer?

A fantastic habit! Once you find $Q$, calculate the vector $\vec{RQ}$ and the direction vector $\vec{d}$. If their dot product is zero, your $Q$ is correct. Then, verify that $Q$ actually lies on the line $AB$ by plugging its coordinates back into the symmetric equation. If both hold, you are golden!

Question

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 $PQ$ is equal to ________

Sigma Percentile · Accepted Answer

### The 3D Canvas: Visualizing the Geometry

My dear student, welcome to the world of 3D geometry. It is a place where intuition meets precision.

Imagine a vast, empty 3D space. Through this space, there runs a straight, infinite line $AB$, defined by two points $A(2, -5, 11)$ and $B(-6, 7, -5)$.

Suspended in this space, like a star in the night sky, is our point $R(1, 7, 6)$. We want to drop a perpendicular from $R$ to the line $AB$. The point where this perpendicular hits the line is $Q$. This is the 'foot' of the perpendicular.

Our mission is to find the distance between this foot $Q$ and another point $P(10, -2, -1)$. Let us begin.

### Phase 1: Defining the Line's Identity

Before we can find $Q$, we must understand the line $AB$. A line is defined by a point and a direction.

We have two points, so we find the direction vector $\vec{d}$ by subtracting the coordinates of $A$ from $B$:
$$\vec{d} = (-6 - 2, 7 - (-5), -5 - 11) = (-8, 12, -16)$$

To make our lives easier, we simplify this vector by dividing by $-4$, giving us $\vec{d} = (2, -3, 4)$. This vector is the 'DNA' of our line; it tells us exactly which way the line is pointing.

### Phase 2: The Parametric Dance

Now, how do we find $Q$? We use the parametric form of the line.

By setting the symmetric equation:
$$\frac{x-2}{2} = \frac{y+5}{-3} = \frac{z-11}{4} = \lambda$$

We introduce $\lambda$, our magic key. Any point on the line can be written as $Q(2\lambda + 2, -3\lambda - 5, 4\lambda + 11)$. As $\lambda$ changes, $Q$ slides along the line.

We need the specific $\lambda$ that makes $RQ$ perpendicular to the line.

### Phase 3: The Perpendicularity Condition

This is the heart of the problem. If $RQ$ is perpendicular to the line, then the vector $\vec{RQ}$ must be orthogonal to the direction vector $\vec{d}$.

Mathematically, this means their dot product is zero: $\vec{RQ} \cdot \vec{d} = 0$. First, we find $\vec{RQ} = Q - R$:
$$\vec{RQ} = (2\lambda + 2 - 1, -3\lambda - 5 - 7, 4\lambda + 11 - 6) = (2\lambda + 1, -3\lambda - 12, 4\lambda + 5)$$

Now, we compute the dot product:
$$2(2\lambda + 1) - 3(-3\lambda - 12) + 4(4\lambda + 5) = 0$$

Expanding this, we get:
$$4\lambda + 2 + 9\lambda + 36 + 16\lambda + 20 = 0$$

Combining terms, $29\lambda + 58 = 0$, which gives us $\lambda = -2$.

### Phase 4: The Final Stretch

With $\lambda = -2$, we find the coordinates of $Q$:
$$x = 2(-2) + 2 = -2$$
$$y = -3(-2) - 5 = 1$$
$$z = 4(-2) + 11 = 3$$

So, $Q = (-2, 1, 3)$. Finally, we calculate the distance $PQ$ between $P(10, -2, -1)$ and $Q(-2, 1, 3)$ using the distance formula:

$$PQ = \sqrt{(10 - (-2))^2 + (-2 - 1)^2 + (-1 - 3)^2}$$
$$PQ = \sqrt{12^2 + (-3)^2 + (-4)^2}$$
$$PQ = \sqrt{144 + 9 + 16} = \sqrt{169} = 13$$

We have arrived at our destination. The length is **13**. You have mastered the geometry of 3D space!

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Direction Ratios of Line $A B$

Equation of Line $A B$

General Coordinates of $Q$

Finding Vector $R Q$

The Perpendicularity Condition

Solving for $λ$

Coordinates of Foot $Q$

The Final Target: Point $P$

Applying Distance Formula

Final Calculation

Conclusion & Summary

The Sigma Insight: Equation of a Line in Space

The 3D Canvas

Phase 1

Phase 2

Phase 3

Phase 4

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