The mirror image of the point (1, 2, 3) in a plane is (-7/3, -4/3, -1/3). Which of the following poi

Q: Wait, why is the line segment $PQ$ the normal vector to the plane?

This is a fundamental property of reflection. When a point is reflected across a plane, the line connecting the point and its image is always perpendicular to the plane. Since the normal vector is defined as any vector perpendicular to the plane, the vector $\vec{PQ}$ (or any scalar multiple of it) serves as our normal vector $\vec{n}$.

Q: Why does the midpoint $M$ of $PQ$ have to lie on the plane?

Think of the plane as the axis of symmetry. The distance from the object to the plane must equal the distance from the image to the plane. Therefore, the plane must pass exactly through the halfway point between the object and the image. That halfway point is the midpoint $M$.

Q: Can I use the point $P$ or $Q$ instead of the midpoint $M$ in the plane equation?

No, you cannot. The points $P$ and $Q$ are on opposite sides of the plane, not on the plane itself. Only the midpoint $M$ is guaranteed to lie on the plane. Always ensure the point you use in the equation $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ is a point that actually lies on the plane.

Question

The mirror image of the point $(1, 2, 3)$ in a plane is $(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3})$. Which of the following points lies on this plane?

Options:

(A) $(1, -1, 1)$

(B) $(-1, -1, 1)$

(C) $(1, 1, 1)$

(D) $(-1, -1, -1)$

Sigma Percentile · Accepted Answer

### The Geometry of Reflection: A Journey into 3D Space

Imagine you are standing in front of a mirror. You see yourself, but you are not actually there; you are in the reflection. In 3D geometry, this is exactly what we are dealing with when we find the mirror image of a point in a plane.

We are given point $P(1, 2, 3)$ and its image $Q(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3})$. Our mission is to find the equation of the plane that acts as this mirror.

### Phase 1: The Normal Vector

The first step is to understand the orientation of the mirror. A plane is defined by its normal vector, $\vec{n}$, which is a vector perpendicular to the plane's surface.

Because the line segment $PQ$ connects an object to its image, it is inherently perpendicular to the mirror plane. Therefore, the vector $\vec{PQ}$ is parallel to the normal vector $\vec{n}$.

We calculate the direction ratios of $\vec{PQ}$ by subtracting the coordinates of $P$ from $Q$:

$$\Delta x = -\frac{7}{3} - 1 = -\frac{10}{3}$$
$$\Delta y = -\frac{4}{3} - 2 = -\frac{10}{3}$$
$$\Delta z = -\frac{1}{3} - 3 = -\frac{10}{3}$$

We now have the direction ratios $(-\frac{10}{3}, -\frac{10}{3}, -\frac{10}{3})$. In vector geometry, we can scale these ratios by any non-zero constant.

To make our calculations simpler, we divide by $-\frac{10}{3}$, which gives us the simplified normal vector $\vec{n} = (1, 1, 1)$. This vector serves as the backbone of our plane equation.

### Phase 2: The Midpoint Anchor

Now that we have the direction, we need a specific point on the plane to anchor our equation. As the mirror is the line of symmetry, the midpoint $M$ of the segment $PQ$ must lie on the plane.

Using the midpoint formula $M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})$, we calculate:

$$x_M = \frac{1 + (-\frac{7}{3})}{2} = -\frac{2}{3}$$
$$y_M = \frac{2 + (-\frac{4}{3})}{2} = \frac{1}{3}$$
$$z_M = \frac{3 + (-\frac{1}{3})}{2} = \frac{4}{3}$$

Our anchor point is $M(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3})$.

### Phase 3: Synthesizing the Equation

With the normal vector $\vec{n} = (1, 1, 1)$ and the point $M(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3})$, we use the point-normal form of the plane equation: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.

Substituting our values, we get:

$$1(x - (-\frac{2}{3})) + 1(y - \frac{1}{3}) + 1(z - \frac{4}{3}) = 0$$

Expanding this, we have:

$$x + \frac{2}{3} + y - \frac{1}{3} + z - \frac{4}{3} = 0$$

Combining the constants, we calculate $\frac{2}{3} - \frac{1}{3} - \frac{4}{3} = -\frac{3}{3} = -1$. Thus, the equation simplifies to $x + y + z - 1 = 0$, or **$x + y + z = 1$**.

### Phase 4: Verification

Finally, we test the point $(1, -1, 1)$ to see if it satisfies the equation $x + y + z = 1$.

Substituting the values:

$$1 + (-1) + 1 = 1$$

The left-hand side equals the right-hand side. The point $(1, -1, 1)$ lies on the plane. You have successfully navigated the 3D space and found the solution.

Animated Solution for Mathematics - Three Dimensional Geometry: The mirror image of the point $(1, 2, 3)$ in a plane is $(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3})$ . Which of the following points lies on this plane?

Select Answer:

Visualized Solution

Visualizing the Points $P$ and $Q$

Properties of a Mirror Image

Direction Ratios of Normal Vector

Simplifying the Normal Vector

Finding the Midpoint $M$

Calculating Midpoint Coordinates

Setting up the Plane Equation

Substituting Values into Plane Equation

Simplifying the Plane Equation

Checking the Options

The Sigma Insight: Equation of a Plane

The Geometry of Reflection

Phase 1

Phase 2

Phase 3

Phase 4

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