Let (α, β, γ) be mirror image of the point (2, 3, 5) in the line (x-1)/2 = (y-2)/3 = (z-3)/4. Then 2

Q: Wait, why do we need the dot product to be zero? Can't we just use distance formula?

Using the distance formula would lead to a system of quadratic equations that is algebraically exhausting! The dot product is the 'geometric shortcut'—it captures the essence of perpendicularity in 3D space without needing to calculate square roots or magnitudes.

Q: How do I know if the point $M$ lies on the line?

Great question! By setting the symmetric equations equal to a parameter $k$, you are essentially defining the coordinates of any arbitrary point on that line. When you solve for $k$ using the perpendicularity condition, you are 'pinpointing' the exact location where the line and the segment $PQ$ intersect.

Q: Is the mirror image always on the same side of the line?

The mirror image is defined by the symmetry across the line. Think of the line as a fold in a piece of paper. If you fold the paper along the line, point $P$ will land exactly on point $Q$. They are equidistant from the line, but on opposite sides of the 'fold'.

Q: Why is the direction vector $(2, 3, 4)$? Where did the $(1, 2, 3)$ go?

The symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ tells us that $(x_0, y_0, z_0)$ is a point on the line, and $(a, b, c)$ is the direction vector. The numbers in the denominator are the components of the direction vector, while the numbers subtracted from $x, y, z$ are just a reference point on the line.

Q: Do I need to find the coordinates of $Q$ individually to solve for $2\alpha + 3\beta + 4\gamma$?

Actually, no! If you look closely at the dot product equation $2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0$, you can see that the expression $2\alpha + 3\beta + 4\gamma$ appears naturally. Sometimes, the JEE rewards you for spotting the algebraic structure before doing the heavy lifting!

Question

Let $(\alpha, \beta, \gamma)$ be mirror image of the point $(2, 3, 5)$ in the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. Then $2\alpha + 3\beta + 4\gamma$ is equal to

Options:

(A) 32

(B) 33

(C) 31

(D) 34

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a point $P(2, 3, 5)$ and a line $L$ defined by the symmetric equations:

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

The line $L$ acts as our mirror. In 3D geometry, the mirror image $Q(\alpha, \beta, \gamma)$ of a point $P$ across a line $L$ implies that the line $L$ is the perpendicular bisector of the segment $PQ$.

### Decoding the Direction

Every line in 3D space is defined by its direction vector. From the symmetric form of the line, we extract the direction ratios from the denominators.

Our direction vector is $\vec{d} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Any line segment perpendicular to our mirror must have a direction vector that, when dotted with $\vec{d}$, yields zero.

### The Power of the Dot Product

We define the vector $\vec{PQ}$ as the displacement from $P$ to $Q$. Mathematically, this is expressed as:

$$\vec{PQ} = (\alpha - 2)\hat{i} + (\beta - 3)\hat{j} + (\gamma - 5)\hat{k}$$

Because the line $L$ is perpendicular to the segment $PQ$, the dot product $\vec{PQ} \cdot \vec{d}$ must vanish:

$$\vec{PQ} \cdot \vec{d} = 0$$

Substituting our components into this condition, we obtain:

$$2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0$$

### The Elegant Conclusion

Now, we expand this expression to isolate the required variables. Distributing the constants yields:

$$2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0$$

Grouping the variables together, we find:

$$2\alpha + 3\beta + 4\gamma - 33 = 0$$

Thus, the expression we were asked to evaluate, $2\alpha + 3\beta + 4\gamma$, is revealed to be exactly **33**.

### Why This Matters

While one could find the coordinates of $Q$ individually by calculating the foot of the perpendicular, the dot product allows us to bypass the individual coordinates entirely. This approach highlights the beauty of JEE-level mathematics: observing the underlying structure of the problem to find the most efficient path to the truth.

You have successfully navigated the mirror, mastering the interplay between vectors and lines. Keep this intuition sharp, as it is a powerful tool for solving complex geometric challenges.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $(α, β, γ)$ be mirror image of the point $(2, 3, 5)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to

Select Answer:

Visualized Solution

The Setup: Point and Line

Defining the Mirror Image

The Geometric Key

Direction Vector of Line $L$

Formulating Vector $P Q$

The Perpendicularity Condition

Applying the Dot Product

Expanding the Equation

Isolating the Target Expression

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Decoding the Direction

The Power of the Dot Product

The Elegant Conclusion

Why This Matters

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