The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, -2,

Q: Wait, the denominator for the x-coordinate is zero. Does that mean the line is undefined or vertical?

That is a fantastic observation! In 3D geometry, a zero in the denominator of a line equation does not mean the line is undefined. It simply means the direction ratio along that axis is zero. It tells us that the line is parallel to the YZ-plane. Specifically, for any point on this line, the x-coordinate is constant. In our case, $x$ is fixed at $1$. It is not a vertical line; it is a line that simply doesn't move in the x-direction.

Q: Why can't we just use the distance formula to minimize the distance between P and the line?

You absolutely could! You could write the distance $d^2 = (x-4)^2 + (y-2)^2 + (z-3)^2$, substitute the parametric forms of $x, y,$ and $z$, and then differentiate with respect to $\lambda$ to find the minimum. However, that involves squaring binomials and solving a quadratic equation, which is algebraically tedious and prone to calculation errors. The dot product method is the 'elegant' path—it is faster, cleaner, and leverages the fundamental property of perpendicularity directly.

Q: What if the dot product wasn't zero? Does that mean the lines aren't perpendicular?

Exactly! The dot product is the litmus test for orthogonality. If $\vec{A} \cdot \vec{B} = 0$, the vectors are perpendicular. If the result is non-zero, the angle between them is not $90^\circ$. In this problem, we are *imposing* the condition that $M$ is the foot of the perpendicular, which forces the dot product to be zero. We are essentially solving for the specific $\lambda$ that satisfies this geometric requirement.

Question

The foot of the perpendicular drawn from the point $(4, 2, 3)$ to the line joining the points $(1, -2, 3)$ and $(1, 1, 0)$ lies on the plane:

Options:

(A) $x - 2y + z = 1$

(B) $x + 2y - z = 1$

(C) $x - y - 2y = 1$

(D) $2x + y - z = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty room. You have a point $P$ floating in the air at coordinates $P(4, 2, 3)$.

Below you, there is a straight line stretching across the room, defined by two points $A(1, -2, 3)$ and $B(1, 1, 0)$. Your task is to drop a plumb line from $P$ to this line.

The point where it lands—the foot of the perpendicular—is $M$. This is the point we must find.

### Phase 1: The Parametric Bridge

The first hurdle in 3D geometry is often the line equation itself. The standard symmetric form of a line passing through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$$\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$$

When we plug in our points $A(1, -2, 3)$ and $B(1, 1, 0)$, we get:

$$\frac{x-1}{1-1} = \frac{y-(-2)}{1-(-2)} = \frac{z-3}{0-3}$$

This simplifies to:

$$\frac{x-1}{0} = \frac{y+2}{3} = \frac{z-3}{-3}$$

Now, set this entire expression equal to a parameter, $\lambda$. This $\lambda$ is your slider; as you vary $\lambda$, you move along the line.

This allows us to express any point $M$ on the line as:

$$M = (1, 3\lambda - 2, -3\lambda + 3)$$

You have just reduced a 3D line to a single variable.

### Phase 2: The Vector Dance

We have our point $P(4, 2, 3)$ and our general point $M(1, 3\lambda - 2, -3\lambda + 3)$. We need to define the vector $\vec{PM}$ that connects them.

To find the direction ratios of this vector, we subtract the coordinates of $P$ from $M$:

$$\vec{PM} = \langle 1-4, (3\lambda-2)-2, (-3\lambda+3)-3 \rangle$$

Simplifying this, we get:

$$\vec{PM} = \langle -3, 3\lambda-4, -3\lambda \rangle$$

This vector represents the 'perpendicular drop' from $P$ to the line.

### Phase 3: The Perpendicularity Constraint

For $M$ to be the foot of the perpendicular, the vector $\vec{PM}$ must be orthogonal to the line $AB$. The direction ratios of the line $AB$ are the denominators found earlier: $\vec{v} = \langle 0, 3, -3 \rangle$.

The condition for orthogonality is that the dot product of the two vectors must be zero:

$$\vec{PM} \cdot \vec{v} = 0$$

Let us perform the calculation:

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

Expanding this, we get:

$$0 + 9\lambda - 12 + 9\lambda = 0$$

This simplifies to $18\lambda = 12$. Solving for $\lambda$, we find:

$$\lambda = \frac{12}{18} = \frac{2}{3}$$

### Phase 4: The Resolution

We substitute $\lambda = \frac{2}{3}$ back into our expression for $M$:

*   The $x$-coordinate remains $1$.
*   The $y$-coordinate is $3(\frac{2}{3}) - 2 = 2 - 2 = 0$.
*   The $z$-coordinate is $-3(\frac{2}{3}) + 3 = -2 + 3 = 1$.

Thus, the foot of the perpendicular is **$M(1, 0, 1)$**.

You have successfully navigated the 3D space, defined the line, applied the orthogonality condition, and arrived at the solution. Remember, geometry is not about memorizing formulas; it is about visualizing the relationship between objects.

Animated Solution for Mathematics - Three Dimensional Geometry: The foot of the perpendicular drawn from the point $(4, 2, 3)$ to the line joining the points $(1, - 2, 3)$ and $(1, 1, 0)$ lies on the plane:

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Visualized Solution

The Geometric Setup

Equation of Line $A B$

Substituting Points $A$ and $B$

Simplified Line Equation

General Point $M$ on the Line

Direction Ratios of $P M$

Simplifying DRs of $P M$

Perpendicularity Condition

Applying the Dot Product

Solving for $λ$

Coordinates of $M$

Checking the Options

The Sigma Insight: Equation of a line in Space

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4