Let P be the image of the point (3, 1, 7) with respect to the plane x - y + z = 3. Then the equation

Q: Wait, why do we use the factor of -2 in the image formula?

That is a fantastic question! Think of the distance from the point to the plane as a vector. To get to the image, you must travel from the point to the plane (that is one distance) and then continue the same distance on the other side. The formula $\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$ gives you the signed distance from the point to the plane. Multiplying by -2 accounts for both the direction (moving across the plane) and the total displacement (twice the distance).

Q: Why can't we just use the midpoint formula to find the image?

You absolutely can! The midpoint of the segment $AP$ must lie on the plane. If you find the line passing through $A$ perpendicular to the plane, you can find the intersection point (the foot of the perpendicular) and then use the midpoint formula to find $P$. However, the formula provided in the hint is just a shortcut that combines these steps into one elegant algebraic expression. Both methods are perfectly valid!

Q: How do I know which point to use for the final plane equation?

The equation of a plane is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. You can use any point that lies on the plane for $(x_0, y_0, z_0)$. In this problem, we have the origin $(0, 0, 0)$ which is on the line $L$, and we have point $P$. Using the origin is mathematically cleaner because it eliminates the constants, making the final equation much simpler to write.

Question

Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x - y + z = 3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$ is

Options:

(A) $x + y - 3z = 0$

(B) $3x + z = 0$

(C) $x - 4y + 7z = 0$

(D) $2x - y = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the mathematical universe! Today, we are going to tackle a problem that involves finding the reflection of a point in a plane and then constructing a new plane from that reflection.

We have a mirror—a plane defined by the equation $x - y + z = 3$. Floating in front of this mirror is a point $A$ with coordinates $(3, 1, 7)$.

Our first mission is to find the image of this point, which we will call $P$. Think of this as finding where the point would appear if you were looking into the mirror.

### Phase 1: The Mirror Reflection

To find the image point $P(x, y, z)$, we rely on a powerful geometric relationship. The line segment connecting $A$ and $P$ must be perpendicular to the plane, and the plane must bisect this segment.

We use the standard formula for the image of a point $(x_1, y_1, z_1)$ in a plane $ax + by + cz + d = 0$:

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

Here, our point $A$ is $(3, 1, 7)$, so $x_1=3, y_1=1, z_1=7$. The plane is $x - y + z - 3 = 0$, so $a=1, b=-1, c=1$, and $d=-3$.

Substituting these values, we get:

$$
\frac{x-3}{1} = \frac{y-1}{-1} = \frac{z-7}{1} = -2 \frac{(3) - (1) + (7) - 3}{1^2 + (-1)^2 + 1^2}
$$

Simplifying the right-hand side, the numerator is $3 - 1 + 7 - 3 = 6$, and the denominator is $1 + 1 + 1 = 3$. Thus, the ratio becomes $-2 	imes (6/3) = -4$.

Now, we solve for $x, y,$ and $z$ individually:
$x-3 = -4 \Rightarrow x = -1$
$y-1 = 4 \Rightarrow y = 5$
$z-7 = -4 \Rightarrow z = 3$

Our image point $P$ is $(-1, 5, 3)$. We have successfully navigated the reflection!

### Phase 2: Building the New Plane

Now, we shift gears. We need to find the equation of a new plane that passes through our newly found point $P(-1, 5, 3)$ and contains the line $L$ given by $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$.

First, observe that the line $L$ passes through the origin $O(0, 0, 0)$. This means our required plane contains both the origin $O$ and the point $P$.

To define a plane, we need a normal vector. We can find this by taking the cross product of two vectors that lie on the plane.

The first vector is the direction vector of the line $L$, which is $\vec{d} = \hat{i} + 2\hat{j} + \hat{k}$. The second vector is the vector connecting the origin to $P$, which is $\vec{OP} = -1\hat{i} + 5\hat{j} + 3\hat{k}$.

The normal vector $\vec{n}$ is the cross product $\vec{d} 	imes \vec{OP}$:

$$
\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ -1 & 5 & 3 \end{vmatrix}
$$

Expanding this determinant:
$\hat{i}(2 \cdot 3 - 1 \cdot 5) - \hat{j}(1 \cdot 3 - 1 \cdot (-1)) + \hat{k}(1 \cdot 5 - 2 \cdot (-1)) = 1\hat{i} - 4\hat{j} + 7\hat{k}$.

Our normal vector is $\vec{n} = \hat{i} - 4\hat{j} + 7\hat{k}$.

### Phase 3: The Final Equation

We have the normal vector $(1, -4, 7)$ and a point on the plane, the origin $(0, 0, 0)$. The general equation of a plane is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.

Substituting our values:
$1(x - 0) - 4(y - 0) + 7(z - 0) = 0$

This simplifies to the final equation:
**$x - 4y + 7z = 0$**

This is the equation of our plane. It is elegant, it is precise, and it perfectly captures the geometry of the problem.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x - y + z = 3$ . Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$ is

Select Answer:

Visualized Solution

Visualizing the Setup

The Image Formula

Substituting the Values

Simplifying the Equation

Calculating Point $P$

The Second Plane and Line

Vectors on the Required Plane

Setting up the Normal Vector

Calculating the Cross Product

Equation of the Plane

Final Substitution

The Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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