Consider the line L given by the equation (x-3)/2 = (y-1)/1 = (z-2)/1. Let Q be the mirror image of

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

Think of the line $L$ as a spear piercing through the plane $P$. If the spear is perpendicular to the plane, then the direction in which the spear is pointing is exactly the same direction as the normal vector (the vector perpendicular to the surface of the plane). In 3D geometry, if a line is perpendicular to a plane, the line's direction vector is parallel to the plane's normal vector. We can simply use the direction ratios of the line as the coefficients of $x, y,$ and $z$ in the plane equation.

Q: Why do we use the parameter $\lambda$ to define point $M$?

The line $L$ is a set of infinite points. We cannot guess which point is the foot of the perpendicular. By setting the equation of the line equal to a parameter $\lambda$, we create a 'scanner' that can represent any point on that line. By imposing the condition that $\vec{AM} \perp L$, we force the math to 'select' the specific $\lambda$ that corresponds to the foot of the perpendicular. It is a powerful algebraic tool to turn a geometric constraint into a solvable equation.

Q: Is there a faster way to find the mirror image $Q$ without finding $M$ first?

While there are advanced vector formulas for reflection, finding the foot of the perpendicular $M$ is the most robust and fundamental method. It builds a deep understanding of the geometry. In JEE Advanced, speed is important, but accuracy is paramount. Mastering the 'find $M$, then find $Q$' workflow ensures you never make a sign error in the reflection formula.

Question

Consider the line $L$ given by the equation $\frac{x-3}{2} = \frac{y-1}{1} = \frac{z-2}{1}$. Let $Q$ be the mirror image of the point $(2, 3, -1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$, and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?

Options:

(A) $(-1, 1, 2)$

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

(C) $(1, 1, 2)$

(D) $(1, 2, 2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The line $L$ is defined by the symmetric equation:

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

We are tasked with finding a plane $P$ such that the line $L$ is perpendicular to $P$ and the point $A(2, 3, -1)$ is reflected across the line $L$ to a point $Q$ that lies on the plane.

### The Normal Vector Epiphany

The orientation of the plane $P$ is determined by its normal vector $\vec{n}$. Since the line $L$ is perpendicular to the plane, the direction vector of the line is parallel to the normal vector of the plane.

From the symmetric form of the line, the direction ratios are $(2, 1, 1)$. Therefore, we define the normal vector as:

$$\vec{n} = 2\hat{i} + \hat{j} + \hat{k}$$

### The Hunt for the Foot of the Perpendicular

To find the mirror image $Q$, we first locate the point $M$ on line $L$ that is closest to $A$. We represent any point $M$ on the line using a parameter $\lambda$:

$$M = (2\lambda+3, \lambda+1, \lambda+2)$$

We construct the vector $\vec{AM}$ by subtracting the coordinates of $A$ from $M$:

$$\vec{AM} = (2\lambda+3-2, \lambda+1-3, \lambda+2-(-1)) = (2\lambda+1, \lambda-2, \lambda+3)$$

### The Dot Product Constraint

Since $M$ is the foot of the perpendicular from $A$ to the line $L$, the vector $\vec{AM}$ must be perpendicular to the direction vector of the line, $\vec{d}_L = (2, 1, 1)$. Thus, their dot product must be zero:

$$\vec{AM} \cdot \vec{d}_L = 0$$

$$(2\lambda+1)(2) + (\lambda-2)(1) + (\lambda+3)(1) = 0$$

Expanding this expression yields:

$$4\lambda + 2 + \lambda - 2 + \lambda + 3 = 0 \Rightarrow 6\lambda + 3 = 0$$

Solving for $\lambda$, we find $\lambda = -\frac{1}{2}$. Substituting this back into the expression for $M$:

$$M = \left(2\left(-\frac{1}{2}\right)+3, -\frac{1}{2}+1, -\frac{1}{2}+2\right) = \left(2, \frac{1}{2}, \frac{3}{2}\right)$$

### The Mirroring Act

For $Q$ to be the mirror image of $A$ across the line, $M$ must be the midpoint of the segment $AQ$. Using the midpoint formula $M = \frac{A+Q}{2}$, we solve for $Q$:

$$Q = 2M - A$$

$$Q = 2\left(2, \frac{1}{2}, \frac{3}{2}\right) - (2, 3, -1) = (4-2, 1-3, 3+1) = (2, -2, 4)$$

### The Final Equation

The plane $P$ passes through $Q(2, -2, 4)$ and has a normal vector $\vec{n} = (2, 1, 1)$. The equation of the plane is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$:

$$2(x-2) + 1(y+2) + 1(z-4) = 0$$

Simplifying the expression:

$$2x - 4 + y + 2 + z - 4 = 0$$

The final equation of the plane is:

$$\mathbf{2x + y + z = 6}$$

Select Answer:

Visualized Solution

Visualizing the 3D Geometry

Identifying the Normal Vector

Defining the General Point $M$

Constructing Vector $A M$

Applying Perpendicularity Condition

Solving for $λ$

Finding Coordinates of $M$

Using the Midpoint Formula

Calculating Mirror Image $Q$

Setting Up the Plane Equation

Simplifying the Plane Equation

Verifying the Options

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal Vector Epiphany

The Hunt for the Foot of the Perpendicular

The Dot Product Constraint

The Mirroring Act

The Final Equation

Similar Questions

Let $Q$ be the mirror image of the point $P (1, 2, 1)$ with respect to the plane $x + 2 y + 2 z = 16$ . Let $T$ be a plane passing through the point $Q$ and contains the line $r = - \hat{k} + λ (\hat{i} + \hat{j} + 2 \hat{k}), λ \in R$ . Then, which of the following points lies on $T$ ?

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?

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?

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The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points ?

Let the lines $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on $P$ ?

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + 4 \hat{k}) = 16$ and $r \cdot (- \hat{i} + \hat{j} + \hat{k}) = 6$ . Then which of the following points does NOT lie on $P$ ?

(i) Find the equation of the plane passing through the points $(2, 1, 0), (5, 0, 1)$ and $(4, 1, 1)$ . (ii) If $P$ is the point $(2, 1, 6)$ then find the point $Q$ such that $P Q$ is perpendicular to the plane in (i) and the midpoint of $P Q$ lies on it.

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The plane passing through the points $(1, 2, 1)$ , $(2, 1, 2)$ and parallel to the line, $2 x = 3 y, z = 1$ also passes through the point:

Select Answer:

Visualized Solution

Visualizing the 3D Geometry

Identifying the Normal Vector

Defining the General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Constructing Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM

Applying Perpendicularity Condition

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

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

Using the Midpoint Formula

Calculating Mirror Image .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Setting Up the Plane Equation

Simplifying the Plane Equation

Verifying the Options

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal Vector Epiphany

The Hunt for the Foot of the Perpendicular

The Dot Product Constraint

The Mirroring Act

The Final Equation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The mirror image of the point (1,2,3) in a plane is (−37​,−34​,−31​). Which of the following points lies on this plane?

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Defining the General Point $M$

Constructing Vector $A M$

Solving for $λ$

Finding Coordinates of $M$

Calculating Mirror Image $Q$

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?

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?

A plane $P$ contains the line $x + 2 y + 3 z + 1 = 0 = x - y - z - 6$ , and is perpendicular to the plane $- 2 x + y + z + 8 = 0$ . Then which of the following points lies on $P$ ?

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points ?

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The plane passing through the points $(1, 2, 1)$ , $(2, 1, 2)$ and parallel to the line, $2 x = 3 y, z = 1$ also passes through the point: