Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a

Q: Wait, why is the direction vector of $PQ$ the same as the normal vector of the plane?

That is a brilliant observation! By definition, the line segment connecting an object to its mirror image is always perpendicular to the mirror surface. Since the normal vector of the plane $S$ is also perpendicular to the surface, the line $PQ$ must be parallel to the normal vector. In our case, the plane is $x + y + z = 5$, so the normal vector is $(1, 1, 1)$. Therefore, the direction ratios of $PQ$ are also $(1, 1, 1)$.

Q: Is there a faster way to find $Q$ without using that long mirror formula?

You could find the foot of the perpendicular (let's call it $M$) first by finding the intersection of the line passing through $P$ with direction $(1, 1, 1)$ and the plane. Then, you could use the midpoint formula, knowing $M$ is the midpoint of $PQ$. However, the mirror formula is essentially a shortcut that combines these steps into one elegant algebraic expression. It is safer and faster for JEE exams!

Q: Why did we set the line $L$ equal to $\lambda$?

Setting the symmetric form of a line equal to a parameter (like $\lambda$) is the most powerful way to handle line-plane intersections. It allows us to represent any arbitrary point on the line as a function of a single variable. When we plug this into the plane equation, we reduce a 3D geometry problem into a simple 1D linear equation in $\lambda$. It turns a complex spatial problem into basic algebra!

Question

Let $Q$ be the mirror image of the point $P(1, 0, 1)$ with respect to the plane $S : x + y + z = 5$. If a line $L$ passing through $(1, -1, -1)$, parallel to the line $PQ$ meets the plane $S$ at $R$, then $QR^2$ is equal to:

Options:

(A) 2

(B) 5

(C) 7

(D) 11

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are constructing a bridge between the abstract world of coordinates and the tangible reality of 3D space.

We are dealing with mirror images, parallel lines, and the precise moment where a line pierces a plane. Take a deep breath, visualize the coordinate axes, and let us begin.

### The Mirror Image

Imagine you are standing in a room. You have a point $P(1, 0, 1)$ floating in space, and there is a mirror, the plane $S: x + y + z = 5$, tilted in front of you. We need to find where the reflection of $P$, which we call $Q$, would appear.

In 3D geometry, the line connecting an object to its reflection is always perpendicular to the mirror. This is the fundamental geometric truth we must anchor ourselves to.

To find the coordinates of $Q(a, b, c)$, we use the powerful mirror image formula:

$$\frac{a-x_1}{A} = \frac{b-y_1}{B} = \frac{c-z_1}{C} = \frac{-2(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2}$$

Here, our plane is $x + y + z - 5 = 0$, so $A=1, B=1, C=1$, and $D=-5$. Our point $P$ is $(1, 0, 1)$. When we substitute these values, the right-hand side of our equation becomes:

$$\frac{-2(1(1) + 1(0) + 1(1) - 5)}{1^2 + 1^2 + 1^2} = \frac{-2(1 + 0 + 1 - 5)}{3} = \frac{-2(-3)}{3} = 2$$

This ratio of $2$ is our key. By equating each coordinate term to this ratio, we find $a-1=2$, $b-0=2$, and $c-1=2$. Thus, our mirror image $Q$ is located at $(3, 2, 3)$.

### The Parallel Path

Now, the problem introduces a new character: a line $L$ passing through $(1, -1, -1)$. We are told this line is parallel to $PQ$.

Let us find the direction of $PQ$. The vector $\vec{PQ}$ is calculated by subtracting the coordinates of $P$ from $Q$: $(3-1, 2-0, 3-1)$, which gives us $(2, 2, 2)$. We can simplify this direction vector to $(1, 1, 1)$.

Since line $L$ is parallel to $PQ$, it must share this same direction. Using the point-direction form of a line, we write the equation of $L$ as:

$$\frac{x-1}{1} = \frac{y+1}{1} = \frac{z+1}{1} = \lambda$$

By setting this equal to a parameter $\lambda$, we can express any point on this line as $R(\lambda + 1, \lambda - 1, \lambda - 1)$. This is the "general point" on our line, waiting for us to find the specific value of $\lambda$ that places it exactly on the plane $S$.

### The Intersection

This is the moment of collision. We need to find where line $L$ pierces the plane $S$. Since point $R$ lies on the plane $x + y + z = 5$, its coordinates must satisfy this equation.

We substitute our general point $R$ into the plane equation:

$$(\lambda + 1) + (\lambda - 1) + (\lambda - 1) = 5$$

Simplifying this, we get $3\lambda - 1 = 5$, which leads to $3\lambda = 6$, or $\lambda = 2$.

With $\lambda = 2$, we can finally pinpoint the exact location of $R$. Substituting $\lambda = 2$ back into our general point expression, we get $R(2+1, 2-1, 2-1)$, which simplifies to $R(3, 1, 1)$.

### Final Calculation

We have arrived at the finish line. We have $Q(3, 2, 3)$ and $R(3, 1, 1)$. The problem asks for the square of the distance between them, $QR^2$.

Using the distance formula:

$$QR^2 = (3 - 3)^2 + (1 - 2)^2 + (1 - 3)^2$$

$$QR^2 = (0)^2 + (-1)^2 + (-2)^2$$

$$QR^2 = 0 + 1 + 4 = 5$$

And there it is. The **final result is 5**.

Look at what we have achieved. We navigated through reflection, vector parallelism, parametric line equations, and 3D distance. You have mastered this problem!

Animated Solution for Mathematics - Three Dimensional Geometry: Let $Q$ be the mirror image of the point $P (1, 0, 1)$ with respect to the plane $S : x + y + z = 5$ . If a line $L$ passing through $(1, - 1, - 1)$ , parallel to the line $P Q$ meets the plane $S$ at $R$ , then $Q R^{2}$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Setup

The Mirror Image Formula

Substituting the Values

Calculating the Ratio

Finding Coordinates of $Q$

Direction of Line $P Q$

Setting up Line $L$

General Point $R$ on Line $L$

Finding Intersection Point $R$

Solving for $λ$ and $R$

Setting up $Q R^{2}$

Final Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Mirror Image

The Parallel Path

The Intersection

Final Calculation

Similar Questions

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A line with positive direction cosines passes through the point $P (2, - 1, 2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x + y + z = 9$ at point $Q$ . The length of the line segment $P Q$ equals

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

Visualizing the Setup

The Mirror Image Formula

Substituting the Values

Calculating the Ratio

Finding Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Direction of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Setting up Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R on Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR2

Final Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Mirror Image

The Parallel Path

The Intersection

Final Calculation

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Finding Coordinates of $Q$

Direction of Line $P Q$

Setting up Line $L$

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