Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T

Q: Wait, why do we use $-2$ in the mirror image formula? Shouldn't it be $-1$ to get to the plane?

That is a brilliant observation! If you use $-1$, you are calculating the foot of the perpendicular—the point where the line from $P$ hits the plane. But the mirror image is on the *other side* of the plane, at the same distance. So, you must travel the distance to the plane *and then* travel that same distance again. Hence, the factor of $-2$.

Q: Can I use any point on the line to form the vector $\vec{AQ}$?

Yes, absolutely! The line $L$ is entirely contained within the plane $T$. This means every single point on that line lies on the plane. You can pick the point given in the vector equation, or any point you get by plugging in a value for $\lambda$. The vector $\vec{AQ}$ will always lie in the plane, and the resulting normal vector will be the same (or a scalar multiple), which is perfectly fine.

Q: Why do we need the cross product? Can't we just use the direction vector of the line as the normal?

Remember, the normal vector must be perpendicular to *every* vector lying in the plane. The direction vector of the line is just one vector in the plane. It is not enough to define the plane's orientation. You need two non-parallel vectors in the plane to define its normal via the cross product. The line's direction vector is only one of them; the vector $\vec{AQ}$ provides the second.

Question

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

Options:

(A) $(2, 1, 0)$

(B) $(1, 2, 1)$

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

(D) $(1, 3, 2)$

Sigma Percentile · Accepted Answer

### The Geometry of Reflections and Planes: A Journey into 3D Space

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through three-dimensional space. In the JEE Advanced arena, 3D geometry is not about memorizing formulas—it is about spatial intuition.

We are going to take a point, reflect it across a mirror, and then construct a plane that holds both this new reflection and a line. Let us break this down, step by step, with the precision of a master architect.

### Phase 1: The Mirror's Secret

Imagine you are standing in front of a mirror. You see your reflection, $Q$, behind the glass. If you are at point $P(1, 2, 1)$ and the mirror is the plane $x + 2y + 2z = 16$, how do we find $Q$?

We use the elegant mirror image formula:

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

Why the $-2$? This is the soul of the formula. If we used $-1$, we would land exactly on the plane—the foot of the perpendicular. But we need to go *through* the plane to the other side, covering the same distance again. So, we double the displacement.

Plugging in our values, $P(1, 2, 1)$ and the plane $x + 2y + 2z - 16 = 0$, we get:

$$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-1}{2} = -2 \frac{1(1) + 2(2) + 2(1) - 16}{1^2 + 2^2 + 2^2}$$

Calculating the right side, the numerator is $1 + 4 + 2 - 16 = -9$, and the denominator is $1 + 4 + 4 = 9$. Thus, the ratio is $-2 	imes (-9 / 9) = 2$.

Equating each coordinate to $2$, we find $x = 3$, $y = 6$, and $z = 5$. Our mirror image $Q$ is at $(3, 6, 5)$. We have successfully anchored our first point in space.

### Phase 2: Constructing the Plane

Now, we enter the second act. We need to find a plane $T$ that contains our point $Q(3, 6, 5)$ and the line $L$ given by $\vec{r} = -\hat{k} + \lambda(\hat{i} + \hat{j} + 2\hat{k})$.

To define a plane, we need a point and a normal vector. We have the point $Q$. We also have a point $A(0, 0, -1)$ on the line $L$. Since both $A$ and $Q$ lie on the plane $T$, the vector $\vec{AQ}$ must also lie on the plane.

$$\vec{AQ} = Q - A = (3-0)\hat{i} + (6-0)\hat{j} + (5 - (-1))\hat{k} = 3\hat{i} + 6\hat{j} + 6\hat{k}$$

We also have the direction vector of the line, $\vec{v} = \hat{i} + \hat{j} + 2\hat{k}$. Now, we have two vectors, $\vec{AQ}$ and $\vec{v}$, both lying flat on the plane. To find the normal vector $\vec{n}$, we take their cross product:

$$\vec{n} = \vec{v} 	imes \vec{AQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 2 \ 3 & 6 & 6 \end{vmatrix}$$

Expanding this determinant, we get $\hat{i}(6 - 12) - \hat{j}(6 - 6) + \hat{k}(6 - 3) = -6\hat{i} + 0\hat{j} + 3\hat{k}$.

### Phase 3: The Final Elegance

We have a normal vector $\vec{n} = -6\hat{i} + 3\hat{k}$. In the world of planes, the magnitude of the normal vector does not change the plane's orientation. We can simplify this by dividing by $-3$, giving us a cleaner normal vector: $\vec{n}_{simplified} = 2\hat{i} - \hat{k}$.

Using the point-normal form $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ with point $A(0, 0, -1)$ and normal $(2, 0, -1)$, we get:

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

$$2x - z - 1 = 0 \Rightarrow 2x - z = 1$$

This is the equation of our plane $T$. Finally, we test our result. When we plug in the point $(1, 2, 1)$, we get $2(1) - 1 = 1$. It satisfies the equation perfectly!

You have navigated the reflection, constructed the plane, and verified the result. This is the power of systematic thinking. Keep this clarity, and no problem will ever be too complex for you. The final equation of the plane is **$2x - z = 1$**.

Select Answer:

Visualized Solution

Visualizing the Problem

The Mirror Image Formula

Substituting Values

Calculating the Ratio

Coordinates of Image $Q$

Introducing Plane $T$ and Line $L$

Extracting Line Information

Vector $A Q$ in Plane $T$

Normal to Plane $T$

Calculating the Cross Product

Simplifying the Normal Vector

Equation of Plane $T$

Final Equation of Plane $T$

Checking the Options

The Sigma Insight: Equation of a Plane

The Geometry of Reflections and Planes

Phase 1

Phase 2

Phase 3

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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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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 $P$ ?

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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 $P$ ?

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(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 :

Select Answer:

Visualized Solution

Visualizing the Problem

The Mirror Image Formula

Substituting Values

Calculating the Ratio

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

Introducing Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T and Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Extracting Line Information

Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AQ​ in Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Normal to Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Calculating the Cross Product

Simplifying the Normal Vector

Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Final Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Checking the Options

The Sigma Insight: Equation of a Plane

The Geometry of Reflections and Planes

Phase 1

Phase 2

Phase 3

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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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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane passing through the point (4, -1, 2) and parallel to the lines 3x+2​=−1y−2​=2z+1​ and 1x−2​=2y−3​=3z−4​ also passes through the point :

Coordinates of Image $Q$

Introducing Plane $T$ and Line $L$

Vector $A Q$ in Plane $T$

Normal to Plane $T$

Equation of Plane $T$

Final Equation of Plane $T$

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