Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2

Q: Wait, why do we use the cross product? Can't we just use the dot product?

Great question! The dot product is used to find the angle between vectors or the projection of one vector onto another. However, here we need a vector that is perpendicular to the entire plane. The cross product is specifically designed to produce a vector that is orthogonal to both input vectors, making it the perfect tool to find the normal vector of a plane.

Q: Why is there a $-2$ in the image formula? It feels arbitrary.

It is not arbitrary at all! The formula $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = k$ represents the line passing through $R$ and perpendicular to the plane. The foot of the perpendicular (let's call it $M$) is where $k = -\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$. Since the image $R'$ is on the other side of the plane at the same distance as $R$, the vector $\vec{RR'}$ is twice the vector $\vec{RM}$. That is why we use $-2$ instead of $-1$.

Q: What if the point $R$ was already on the plane?

If $R$ is on the plane, then $ax_1 + by_1 + cz_1 + d = 0$. If you plug this into the formula, the right-hand side becomes zero. This means $x = x_1, y = y_1, z = z_1$. In other words, the image of a point on the plane is the point itself. The math is consistent!

Question

Let $P$ be a plane passing through the points $(2, 1, 0), (4, 1, 1)$ and $(5, 0, 1)$ and $R$ be any point $(2, 1, 6)$. Then the image of $R$ in the plane $P$ is:

Options:

(A) $(6, 5, 2)$

(B) $(6, 5, -2)$

(C) $(4, 3, 2)$

(D) $(3, 4, -2)$

Sigma Percentile · Accepted Answer

### The Geometry of Reflection: Finding the Image in a Plane

Imagine you are standing in a vast, three-dimensional void. Before you floats a perfectly flat, infinite sheet of paper—a plane $P$.

You are given three specific coordinates on this plane: $A(2, 1, 0)$, $B(4, 1, 1)$, and $C(5, 0, 1)$. These three points are the anchors that define the orientation of our plane.

Hovering above this plane is a point $R(2, 1, 6)$. Your task is to find the reflection of $R$ in this mirror-like plane. This is not just a calculation; it is a journey into the heart of spatial geometry.

### Phase 1: Defining the Mirror

Before we can find the reflection, we must define the mirror itself. A plane is defined by its normal vector—a vector that stands perfectly perpendicular to its surface.

To find this normal vector $\vec{n}$, we need two vectors that lie flat on the plane. We can easily create these by connecting our given points.

First, let us find vector $\vec{AB}$ by subtracting the position vector of $A$ from $B$: 
$$\vec{AB} = (4-2)\hat{i} + (1-1)\hat{j} + (1-0)\hat{k} = 2\hat{i} + 0\hat{j} + 1\hat{k}$$

Next, we find vector $\vec{AC}$ by subtracting $A$ from $C$: 
$$\vec{AC} = (5-2)\hat{i} + (0-1)\hat{j} + (1-0)\hat{k} = 3\hat{i} - 1\hat{j} + 1\hat{k}$$

Now, we invoke the power of the cross product. The cross product of $\vec{AB}$ and $\vec{AC}$ gives us a vector that is perpendicular to both, and thus perpendicular to the plane. We compute this using a determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & 1 \ 3 & -1 & 1 \end{vmatrix}$$

Expanding this, we get $\vec{n} = \hat{i}(0 - (-1)) - \hat{j}(2 - 3) + \hat{k}(-2 - 0) = \hat{i} + \hat{j} - 2\hat{k}$. Our normal vector is $(1, 1, -2)$.

### Phase 2: The Equation of the Plane

With the normal vector $(a, b, c) = (1, 1, -2)$ and a point on the plane $A(2, 1, 0)$, we can write the equation of the plane using the point-normal form: $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.

Substituting our values, we get:

$$1(x - 2) + 1(y - 1) - 2(z - 0) = 0$$

$$x - 2 + y - 1 - 2z = 0$$

$$x + y - 2z - 3 = 0$$

This is the mathematical identity of our mirror. It tells us exactly how the plane is tilted and positioned in space.

### Phase 3: The Mirror Image

Now, we reach the climax of our journey. We need the image $R'(x, y, z)$ of point $R(2, 1, 6)$. We use the elegant image formula, which relates the coordinates of the image to the original point and the plane coefficients:

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

Let us calculate the right-hand side constant. Substituting $R(2, 1, 6)$ into the plane equation $x + y - 2z - 3 = 0$ for the numerator, and the sum of squares of the normal vector components for the denominator:

$$	ext{Numerator} = 1(2) + 1(1) - 2(6) - 3 = 2 + 1 - 12 - 3 = -12$$

$$	ext{Denominator} = 1^2 + 1^2 + (-2)^2 = 1 + 1 + 4 = 6$$

$$	ext{RHS} = -2 \left( \frac{-12}{6} ight) = -2(-2) = 4$$

Now, we simply equate each coordinate expression to $4$:

1. $\frac{x - 2}{1} = 4 \Rightarrow x - 2 = 4 \Rightarrow x = 6$

2. $\frac{y - 1}{1} = 4 \Rightarrow y - 1 = 4 \Rightarrow y = 5$

3. $\frac{z - 6}{-2} = 4 \Rightarrow z - 6 = -8 \Rightarrow z = -2$

And there it is! The image of point $R$ in the plane $P$ is **$R'(6, 5, -2)$**. We have successfully navigated the 3D space, defined the plane, and calculated the reflection.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $P$ be a plane passing through the points $(2, 1, 0), (4, 1, 1)$ and $(5, 0, 1)$ and $R$ be any point $(2, 1, 6)$ . Then the image of $R$ in the plane $P$ is:

Select Answer:

Visualized Solution

Visualizing the Plane

The Objective: Finding the Image

Strategy: Equation of the Plane

Calculating Vector $A B$

Calculating Vector $A C$

The Normal Vector $n$

Computing the Cross Product

Setting up the Plane Equation

Finalizing the Plane Equation

The Image Formula

Substituting into the Formula

Evaluating the Right Hand Side

Solving for $x$ and $y$

Solving for $z$

The Final Image Point

The Sigma Insight: Equation of a Plane

The Geometry of Reflection

Phase 1

Phase 2

Phase 3

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