The image of the line (x-1)/3 = (y-3)/1 = (z-4)/-5 in the plane 2x - y + z + 3 = 0 is the line: (a)

Q: Why do we check the dot product $\vec{d} \cdot \vec{n}$ first?

It is the litmus test for the relationship between the line and the plane. If the dot product is zero, the line is parallel to the plane. If it were non-zero, the line would intersect the plane, and the image would be a different line passing through that intersection point. Always check this first to save time and avoid using the wrong strategy!

Q: Why does the image line have the same direction ratios as the original?

Think of the plane as a mirror. When you place a stick parallel to a mirror, its reflection is also parallel to the mirror. Since the direction ratios define the orientation of the line, and reflection in a plane preserves the angle of the line relative to the plane, the orientation remains unchanged. It is a beautiful geometric symmetry.

Q: What does the $-2$ in the point-image formula actually represent?

The formula $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = k$ gives the coordinates of the foot of the perpendicular when $k = -\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$. By using $-2$ instead of $-1$, we are effectively jumping twice the distance from the point to the plane, which lands us exactly on the other side—the image point.

Question

The image of the line $\frac{x-1}{3} = \frac{y-3}{1} = \frac{z-4}{-5}$ in the plane $2x - y + z + 3 = 0$ is the line:

Options:

(A) $\frac{x-3}{3} = \frac{y+5}{1} = \frac{z-2}{-5}$

(B) $\frac{x-3}{-3} = \frac{y+5}{-1} = \frac{z-2}{5}$

(C) $\frac{x+3}{3} = \frac{y-5}{1} = \frac{z-2}{-5}$

(D) $\frac{x+3}{-3} = \frac{y-5}{-1} = \frac{z+2}{5}$

Sigma Percentile · Accepted Answer

### The Dance of Geometry: Reflecting a Line in a Plane

Welcome, future engineers. Today, we are not just solving a problem; we are visualizing the elegant dance of 3D geometry. We are tasked with finding the image of a line, $L$, in a plane, $\Pi$.

Imagine you are standing in a room, holding a laser pointer parallel to the floor. If the floor is a mirror, where does the reflection of that laser beam appear? That is the physical reality behind this algebraic challenge.

### Phase 1: The Anatomy of the Line and Plane

First, let us extract the DNA of our objects. From the symmetric form of the line:

$$\frac{x-1}{3} = \frac{y-3}{1} = \frac{z-4}{-5}$$

We immediately identify a point $A(1, 3, 4)$ and the direction vector $\vec{d} = (3, 1, -5)$. These are the coordinates and the orientation of our laser beam.

Next, we look at the plane $2x - y + z + 3 = 0$. The coefficients of $x, y,$ and $z$ give us the normal vector $\vec{n} = (2, -1, 1)$, which points straight out of the plane like a spear.

### Phase 2: The Parallelism Test

Before we rush into complex reflections, we must ask: are the line and the plane parallel? If they intersect, the strategy changes entirely.

We perform the dot product test: $\vec{d} \cdot \vec{n} = (3)(2) + (1)(-1) + (-5)(1)$. Calculating this, we get $6 - 1 - 5 = 0$.

The dot product is zero! This is a moment of triumph. It tells us that the line is perfectly parallel to the plane. Because the line is parallel to the plane, its reflection will also be parallel to the plane, sharing the exact same direction ratios $(3, 1, -5)$.

We have just simplified our problem significantly; we only need to find the image of one point on the line to define the entire image line.

### Phase 3: The Reflection

We choose our point $A(1, 3, 4)$ and prepare to reflect it across the plane. We use the standard point-image formula:

$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$$

This formula is a powerful tool in your JEE arsenal. It calculates the displacement required to jump from the original point, through the plane, to the image point $P(x, y, z)$.

Substituting our values, we get:

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

Look at the numerator: $2(1) - 3 + 4 + 3 = 6$. Look at the denominator: $4 + 1 + 1 = 6$.

The ratio is $6/6 = 1$. Multiplying by the $-2$ outside, our scaling factor is $-2$. This is the beauty of well-constructed problems—the numbers often collapse into simplicity if you trust the process.

### Phase 4: The Final Construction

Now, we solve for the coordinates of $P$:

$x - 1 = 2(-2) \Rightarrow x = -3$

$y - 3 = -1(-2) \Rightarrow y = 5$

$z - 4 = 1(-2) \Rightarrow z = 2$

Our image point is $P(-3, 5, 2)$. We have the point, and we have the direction ratios $(3, 1, -5)$.

Putting it all together into the symmetric form, we get the final equation of the reflected line:

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

You have successfully navigated the 3D space. Remember, geometry is not about memorizing formulas; it is about understanding the relationship between objects.

When you see a line and a plane, always check their orientation first. Once you do that, the path to the solution becomes clear. Keep practicing, keep visualizing, and keep falling in love with the logic of physics and math.

Animated Solution for Mathematics - Three Dimensional Geometry: The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane $2 x - y + z + 3 = 0$ is the line:

Select Answer:

Visualized Solution

Identifying the Line and Plane

Checking for Parallelism

Strategy for the Image Line

Image of Point $A (1, 3, 4)$

Calculating the Scaling Factor

Finding the Image Point $P$

The Final Image Line Equation

The Sigma Insight: Intersection of a Line and a Plane

The Dance of Geometry

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The image of the line 3x−1​=1y−3​=−5z−4​ in the plane 2x−y+z+3=0 is the line: