Let the foot of the perpendicular from the point (1, 2, 4) on the line (x+2)/4 = (y-1)/2 = (z+1)/3 b

Q: Why do we introduce the parameter $\lambda$ in the first place?

Think of $\lambda$ as a 'slider' or a coordinate system for the line itself. Since a line in 3D space has infinitely many points, we cannot describe it with a single coordinate. By setting the ratios equal to $\lambda$, we create a bridge that allows us to express any point $(x, y, z)$ on that line as a function of just one variable. This turns a complex 3D geometry problem into a simple 1D algebraic equation.

Q: Why is the dot product of $\vec{AP}$ and the direction vector $\vec{d}$ equal to zero?

This is the geometric definition of perpendicularity. If two vectors are perpendicular, the angle between them is $90^\circ$. Since the dot product is defined as $\vec{A} \cdot \vec{B} = |A||B| \cos(\theta)$, and $\cos(90^\circ) = 0$, the dot product must be zero. It is the most powerful tool in your arsenal for enforcing orthogonality in 3D space.

Q: What if the distance formula gives me a negative value?

Distance is a physical quantity and must always be non-negative. The formula includes an absolute value sign $|ax_1 + by_1 + cz_1 + d|$ precisely for this reason. If you calculate the expression inside and get a negative number, just take its magnitude. It simply means the point lies on the 'negative' side of the plane relative to the normal vector.

Question

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3}$ be $P$. Then the distance of $P$ from the plane $3x + 4y + 12z + 23 = 0$

Options:

(A) 5

(B) $\frac{50}{13}$

(C) 4

(D) $\frac{63}{13}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. Before you, there is a fixed point $A(1, 2, 4)$ and a line stretching infinitely in both directions, defined by the equation:

$$\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3}$$

Your mission is to find the exact point $P$ on this line that is closest to $A$. This is the 'foot of the perpendicular,' a fundamental building block for understanding projections and the structure of space.

### Phase 1: The Parametric Key

To find $P$, we need a way to label every point on the line. We use a parameter, let us call it $\lambda$. By setting the line equation equal to $\lambda$, we unlock the ability to write $x, y,$ and $z$ as functions of $\lambda$:

$$x = 4\lambda - 2, \quad y = 2\lambda + 1, \quad z = 3\lambda - 1$$

Any point on this line is just a specific value of $\lambda$ away. Our point $P$ is simply one of these points, waiting to be discovered.

### Phase 2: The Language of Orthogonality

We know that the vector $\vec{AP}$ must be perpendicular to the line. The line itself has a direction vector $\vec{d} = (4, 2, 3)$, which we extract directly from the denominators of the line equation.

The vector $\vec{AP}$ is found by subtracting the coordinates of $A$ from our parametric coordinates of $P$:

$$\vec{AP} = (4\lambda - 3, 2\lambda - 1, 3\lambda - 5)$$

Because $\vec{AP}$ is perpendicular to the line, their dot product must vanish: $\vec{AP} \cdot \vec{d} = 0$. We compute the dot product:

$$4(4\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 5) = 0$$

Expanding this, we get $16\lambda - 12 + 4\lambda - 2 + 9\lambda - 15 = 0$. Combining terms, we arrive at $29\lambda - 29 = 0$, which yields the elegant result $\lambda = 1$.

Substituting $\lambda = 1$ back into our parametric equations, we find the coordinates of $P$:

$$x = 4(1) - 2 = 2, \quad y = 2(1) + 1 = 3, \quad z = 3(1) - 1 = 2$$

Thus, the foot of the perpendicular is **$P(2, 3, 2)$**.

### Phase 3: The Final Leap to the Plane

Now that we have pinned down $P(2, 3, 2)$, we are given a plane $3x + 4y + 12z + 23 = 0$ and asked for the perpendicular distance from $P$ to this plane. We invoke the standard distance formula:

$$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Plugging in our values, we get:

$$d = \frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{3^2 + 4^2 + 12^2}}$$

The numerator becomes $|6 + 12 + 24 + 23| = |65| = 65$. The denominator is $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

Finally, the perpendicular distance is:

$$d = \frac{65}{13} = 5$$

The final result is **$5$**. Remember, in JEE Advanced, complexity is often just a mask for simple, elegant principles. Keep visualizing, keep calculating, and never lose sight of the geometry.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}$ be $P$ . Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$

Select Answer:

Visualized Solution

Visualizing the Setup

Parametric Form of Point $P$

Finding Vector $A P$

The Perpendicularity Condition

Setting up the Equation

Solving for $λ$

Coordinates of Foot $P$

Introducing the Plane

Distance Formula Setup

Final Calculation

Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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