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 normal vector parallel to the vector $\vec{PQ}$?

Think about the geometry of reflection. The line segment connecting a point to its image in a plane is always perpendicular to that plane. Since the vector $\vec{PQ}$ lies along this line, its direction is identical to the direction of the plane's normal vector. Therefore, we can use the components of $\vec{PQ}$ directly as the direction ratios of the normal.

Q: Do I really need to find the full equation of the plane? Can't I just plug the points into the options?

You could potentially test the points, but you would need to know the plane's equation first to verify which point satisfies it! The most robust way is to derive the equation $x + y + z = 1$ using the perpendicular bisector property. Once you have that, checking the options becomes a simple arithmetic task.

Q: I'm getting confused with the fractions. Is there a trick to simplify the normal vector?

Absolutely! When you calculate $\vec{PQ}$, you will get components like $-\frac{10}{3}$. Remember, direction ratios are proportional. You don't need to use the exact values $-\frac{10}{3}, -\frac{10}{3}, -\frac{10}{3}$. You can divide the entire vector by the common factor $-\frac{10}{3}$ to get the simplest direction ratios: $(1, 1, 1)$. This makes the final equation much easier to handle.

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 Mirror in 3D Space

Imagine you are standing in front of a mirror. You see your reflection, a perfect twin located exactly behind the glass. In the world of 3D geometry, this is not just a visual trick; it is a precise mathematical relationship.

When we talk about the mirror image of a point $P$ in a plane, we are essentially saying that the plane is the perpendicular bisector of the line segment connecting $P$ and its image $Q$. This is the key to unlocking this problem.

### Phase 1: The Vector Bridge

To define a plane, we need two fundamental pieces of information: a point that lies on the plane and a normal vector that defines its orientation. Our problem gives us the point $P(1, 2, 3)$ and its image $Q(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3})$.

Because the plane is the perpendicular bisector of the segment $PQ$, the vector $\vec{PQ}$ is inherently perpendicular to the plane. This makes $\vec{PQ}$ our perfect candidate for the normal vector $\vec{n}$.

Let us calculate it:
$$\vec{PQ} = Q - P = \left(-\frac{7}{3} - 1\right)\hat{i} + \left(-\frac{4}{3} - 2\right)\hat{j} + \left(-\frac{1}{3} - 3\right)\hat{k}$$

Simplifying this, we get:
$$\vec{PQ} = \left(-\frac{10}{3}\right)\hat{i} + \left(-\frac{10}{3}\right)\hat{j} + \left(-\frac{10}{3}\right)\hat{k}$$

Since we only care about the direction of the normal, we can scale this vector by dividing by $-\frac{10}{3}$, giving us the beautifully simple normal vector $\vec{n} = (1, 1, 1)$.

### Phase 2: The Anchor Point

Now that we have our normal vector, we need an anchor—a point that we know for certain lies on the plane. Since the plane bisects the segment $PQ$, the midpoint $M$ of $PQ$ must lie on the plane.

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

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

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

### Phase 3: Constructing the Plane

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

Substituting our values, we get:
$$1\left(x - \left(-\frac{2}{3}\right)\right) + 1\left(y - \frac{1}{3}\right) + 1\left(z - \frac{4}{3}\right) = 0$$

Expanding this, we have:
$$x + \frac{2}{3} + y - \frac{1}{3} + z - \frac{4}{3} = 0$$

Combining the constants, we get:
$$x + y + z + \frac{2 - 1 - 4}{3} = 0$$

This simplifies to $x + y + z - 1 = 0$, or **$x + y + z = 1$**.

### Phase 4: The Final Verification

We have arrived at the equation of the plane: $x + y + z = 1$. Now, we simply check the given options to see which point satisfies this condition.

Testing the point $(1, -1, 1)$, we find:
$$1 + (-1) + 1 = 1$$

It satisfies the equation perfectly! This journey through the geometry of reflection shows us that even complex 3D problems can be broken down into simple, elegant steps.

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

The Mirror Image Setup

Normal Vector to the Plane

Calculating Vector $P Q$

Direction Ratios of Normal

Finding a Point on the Plane

Applying Midpoint Formula

Coordinates of Midpoint $M$

Equation of the Plane

Substituting Values into Plane Equation

Simplifying the Equation

Verifying the Given Options

The Sigma Insight: Equation of a Plane

The Geometry of Reflection

Phase 1

Phase 2

Phase 3

Phase 4

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