If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1

Q: Wait, why do we use the cross product of the normals? Shouldn't we use the planes themselves?

That is a common point of confusion! Remember, a plane is defined by its normal vector. If a line is parallel to a plane, it means the line is perpendicular to the normal of that plane. Since the line is parallel to *two* planes, it must be perpendicular to *both* normals. The cross product is the mathematical tool that gives us a vector perpendicular to two given vectors.

Q: Why do we introduce $\lambda$?

Think of $\lambda$ as a 'position marker' along the line. By setting the line equation equal to $\lambda$, we can express any point $M$ on the line as a function of a single variable. This turns a 3D geometry problem into a simple 1D algebraic equation, which is much easier to solve!

Q: Is there a faster way than the dot product?

While there are other methods involving projection, the dot product method is the most robust and 'JEE-proof' technique. It directly enforces the geometric definition of perpendicularity, minimizing the chance of calculation errors. Stick with the dot product—it is your best friend in 3D geometry.

Question

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 $x + 2y + z = 0$ and $3y - z = 3$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to

Options:

(A) -1

(B) 3

(C) 1

(D) 5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have a point $P(1, 9, 7)$ floating in the air, and a line passing through point $A(3, 2, 1)$.

Your goal is to find the 'foot of the perpendicular'—the exact point $M$ on that line that is closest to $P$. This is a quest to find the shortest path from a point to a line in 3D space.

### Finding the Direction

The line is defined by its relationship to two planes: $x + 2y + z = 0$ and $3y - z = 3$. Since the line is parallel to both planes, it must be perpendicular to the normal vectors of both planes.

The normal vectors are $\vec{n}_1 = (1, 2, 1)$ and $\vec{n}_2 = (0, 3, -1)$. We find the direction vector $\vec{b}$ of the line by calculating the cross product $\vec{b} = \vec{n}_1 	imes \vec{n}_2$:

$$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 0 & 3 & -1 \end{vmatrix}$$

Expanding this determinant, we get:

$$\vec{b} = \hat{i}(-2 - 3) - \hat{j}(-1 - 0) + \hat{k}(3 - 0) = -5\hat{i} + \hat{j} + 3\hat{k}$$

Thus, the direction vector is $\vec{b} = (-5, 1, 3)$.

### The Parametric Path

Given the point $A(3, 2, 1)$ and the direction vector $\vec{b} = (-5, 1, 3)$, the equation of the line in symmetric form is:

$$\frac{x-3}{-5} = \frac{y-2}{1} = \frac{z-1}{3} = \lambda$$

This allows us to represent any arbitrary point $M$ on the line using the parameter $\lambda$:

$$M = (-5\lambda + 3, \lambda + 2, 3\lambda + 1)$$

### The Perpendicularity Condition

We seek the point $M$ such that the vector $\vec{PM}$ is perpendicular to the line. First, we calculate $\vec{PM} = M - P$:

$$\vec{PM} = (-5\lambda + 3 - 1, \lambda + 2 - 9, 3\lambda + 1 - 7) = (-5\lambda + 2, \lambda - 7, 3\lambda - 6)$$

Since $\vec{PM}$ is perpendicular to the line, its dot product with the direction vector $\vec{b}$ must be zero:

$$\vec{PM} \cdot \vec{b} = -5(-5\lambda + 2) + 1(\lambda - 7) + 3(3\lambda - 6) = 0$$

Expanding the terms, we obtain:

$$25\lambda - 10 + \lambda - 7 + 9\lambda - 18 = 0$$

$$35\lambda - 35 = 0 \Rightarrow \lambda = 1$$

### Final Calculation

Substituting $\lambda = 1$ back into our expression for $M$, we find the coordinates $(\alpha, \beta, \gamma)$:

$$\alpha = -5(1) + 3 = -2$$
$$\beta = 1 + 2 = 3$$
$$\gamma = 3(1) + 1 = 4$$

The foot of the perpendicular is $M(-2, 3, 4)$. The sum of these coordinates is:

$$\alpha + \beta + \gamma = -2 + 3 + 4 = \mathbf{5}$$

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Select Answer:

Visualized Solution

Visualizing the Geometry

Direction Vector of the Line

Setting up the Cross Product

Calculating $b$

Equation of the Line

Coordinates of Foot $M$

Vector $P M$

Applying Perpendicularity Condition

Setting up the Dot Product

Solving for $λ$

Finding the Foot Coordinates

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Direction

The Parametric Path

The Perpendicularity Condition

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Geometry

Direction Vector of the Line

Setting up the Cross Product

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Equation of the Line

Coordinates of Foot .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PM

Applying Perpendicularity Condition

Setting up the Dot Product

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Foot Coordinates

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Direction

The Parametric Path

The Perpendicularity Condition

Final Calculation

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