If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is Q(α

Q: Why do we need to find the foot of the perpendicular $R$ first? Can't we just use reflection formulas directly?

While there are direct reflection formulas, they are often prone to sign errors. Finding the foot of the perpendicular $R$ is a robust, step-by-step method that builds a deeper geometric intuition. It ensures you understand that the image is simply the point reflected across the line, maintaining the symmetry of the perpendicular bisector.

Q: How do I know which direction vector to use for the line $AB$?

The direction vector of a line passing through two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is simply the vector $\vec{AB} = (x_2-x_1, y_2-y_1, z_2-z_1)$. Any scalar multiple of this vector is also a valid direction vector, so don't worry if your signs are flipped; the dot product will still yield the same result!

Q: I got $\lambda = -4/3$. Does the negative sign mean I did something wrong?

Absolutely not! The parameter $\lambda$ is just a coordinate marker along the line. A negative value simply means that the point $R$ lies in the direction opposite to the vector $\vec{AB}$ starting from your chosen reference point. It is perfectly valid.

Q: Why is the dot product of $\vec{PR}$ and the line's direction vector equal to zero?

Because $R$ is the foot of the perpendicular from $P$ to the line $AB$. By definition, the segment $PR$ must be perpendicular to the line itself. In vector algebra, two vectors are perpendicular if and only if their dot product is zero. This is the most powerful tool in your 3D geometry arsenal!

Q: Is there a shortcut to find $\alpha + \beta + \gamma$ without finding each coordinate individually?

While you could potentially manipulate the vector equations, finding the individual coordinates is the safest path to avoid algebraic traps. In JEE Advanced, precision is your greatest asset. Calculating $\alpha, \beta, \gamma$ individually ensures you don't miss a hidden constant or a sign error.

Question

If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma$ is equal to

Options:

(A) $47/3$

(B) $46/3$

(C) $18$

(D) $13$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You have a point $P(1, 0, 3)$ floating in space, and a line $AB$ defined by the points $A(4, 7, 1)$ and $B(3, 5, 3)$.

You are tasked with finding the 'image' of $P$ in this line. Think of this line as a mirror. If you were to look into this mirror, where would the reflection $Q(\alpha, \beta, \gamma)$ appear?

### Defining the Mirror

Before we can find the reflection, we must understand the mirror itself. A line in 3D space is defined by a point and a direction.

The direction ratios of our line are the differences in the coordinates of $A$ and $B$:
$$\vec{d} = (4-3, 7-5, 1-3) = (1, 2, -2)$$

This vector acts as the compass for our line. We can now write the equation of the line in its symmetric form:
$$\frac{x-3}{1} = \frac{y-5}{2} = \frac{z-3}{-2} = \lambda$$

This parameter $\lambda$ is our key. It allows us to represent any point $R$ on the line as a function of a single variable:
$$R = (\lambda+3, 2\lambda+5, -2\lambda+3)$$

### The Perpendicular Connection

The line segment connecting the original point $P$ to its image $Q$ must be perpendicular to the mirror line $AB$. Furthermore, the point of intersection $R$ must be the midpoint of $PQ$.

To find $R$, we look for the specific value of $\lambda$ that makes the vector $\vec{PR}$ perpendicular to the line's direction vector $\vec{d}$. We calculate $\vec{PR} = R - P$:
$$\vec{PR} = ((\lambda+3)-1, (2\lambda+5)-0, (-2\lambda+3)-3) = (\lambda+2, 2\lambda+5, -2\lambda)$$

For $\vec{PR}$ to be perpendicular to the line, their dot product must vanish:
$$\vec{PR} \cdot \vec{d} = 0$$

Substituting our values, we get:
$$(\lambda+2)(1) + (2\lambda+5)(2) + (-2\lambda)(-2) = 0$$

Expanding this, we find:
$$\lambda + 2 + 4\lambda + 10 + 4\lambda = 0$$

This simplifies to $9\lambda + 12 = 0$, which yields the parameter:
$$\lambda = -\frac{4}{3}$$

### The Final Symmetry

With $\lambda = -\frac{4}{3}$, we locate the foot of the perpendicular $R$:
$$x_R = -\frac{4}{3} + 3 = \frac{5}{3}, \quad y_R = 2(-\frac{4}{3}) + 5 = \frac{7}{3}, \quad z_R = -2(-\frac{4}{3}) + 3 = \frac{17}{3}$$

Now, we invoke the property of reflection: $R$ is the midpoint of $PQ$. Thus, $Q = 2R - P$. Calculating the coordinates of $Q$:
$$\alpha = 2\left(\frac{5}{3}\right) - 1 = \frac{7}{3}, \quad \beta = 2\left(\frac{7}{3}\right) - 0 = \frac{14}{3}, \quad \gamma = 2\left(\frac{17}{3}\right) - 3 = \frac{25}{3}$$

Finally, we sum these values to find our **final answer**:
$$\alpha + \beta + \gamma = \frac{7}{3} + \frac{14}{3} + \frac{25}{3} = \frac{46}{3}$$

Animated Solution for Mathematics - Three Dimensional Geometry: If the image of the point $P (1, 0, 3)$ in the line joining the points $A (4, 7, 1)$ and $B (3, 5, 3)$ is $Q (α, β, γ)$ , then $α + β + γ$ is equal to

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Visualized Solution

Visualizing the Geometry

Direction Ratios of Line $A B$

Equation of Line $A B$

General Point $R$ on the Line

Vector $P R$ Calculation

Condition for Perpendicularity

Solving for $λ$ - Expansion

Solving for $λ$ - Final Value

Coordinates of Foot $R$

Finding the Image $Q$

Final Sum $α + β + γ$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Defining the Mirror

The Perpendicular Connection

The Final Symmetry

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Visualized Solution

Visualizing the Geometry

Direction Ratios of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R on the Line

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

Condition for Perpendicularity

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

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

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

Finding the Image .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Final Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+β+γ