A plane which is perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z = 4, passes through (1,

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

Great question! The dot product is used to check if two vectors are perpendicular (if their dot product is zero). However, the cross product is the tool we use to *construct* a vector that is simultaneously perpendicular to two given vectors. Since our required plane's normal must be perpendicular to both given normals, the cross product is the direct path to finding that direction.

Q: Why did we scale the normal vector $\vec{n} = (-3, -3, 0)$ to $(1, 1, 0)$? Does it change the plane?

Not at all! The equation of a plane $Ax + By + Cz + D = 0$ is defined by the *direction* of the normal, not its magnitude. Scaling the normal vector by a constant (like dividing by $-3$) simply changes the coefficients $A, B, C$ proportionally, which results in an equivalent equation. It makes your arithmetic much cleaner and reduces the chance of calculation errors.

Q: Why does the $z$ term disappear in the final plane equation?

That is a beautiful observation! When the $k$-component of the normal vector is zero, it means the normal vector lies entirely in the $xy$-plane. Geometrically, this implies that the plane itself is parallel to the $z$-axis. It's not a mistake; it's a specific geometric property of the plane we found.

Question

A plane which is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$, passes through $(1, -2, 1)$. The distance of the plane from the point $(1, 2, 2)$ is

Options:

(A) $0$

(B) $1$

(C) $\sqrt{2}$

(D) $2\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast 3D coordinate system. You are given two planes, $P_1$ and $P_2$, and you are tasked with finding a third plane that is perpendicular to both.

This is a classic JEE Advanced challenge that tests your ability to visualize vectors in space. Let us break this down step by step.

### Extracting the Normals

Every plane has a soul, and that soul is its normal vector. For the first plane, $2x - 2y + z = 0$, the coefficients of $x, y,$ and $z$ reveal the normal vector $\vec{n}_1 = (2, -2, 1)$.

Similarly, for the second plane, $x - y + 2z = 4$, the normal vector is $\vec{n}_2 = (1, -1, 2)$. These vectors are the compasses of our planes; they tell us exactly which way the planes are facing.

### The Magic of the Cross Product

We need a new plane that is perpendicular to both $P_1$ and $P_2$. Geometrically, this means the normal vector of our new plane, let us call it $\vec{n}$, must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$.

This is the perfect moment to deploy the cross product. The cross product $\vec{n} = \vec{n}_1 	imes \vec{n}_2$ is the mathematical key that unlocks a vector perpendicular to two others. We set up the determinant:

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

Expanding this carefully, we calculate the components:
The $\hat{i}$ component is $(-2)(2) - (1)(-1) = -4 + 1 = -3$.
The $\hat{j}$ component is $-((2)(2) - (1)(1)) = -(4 - 1) = -3$.
The $\hat{k}$ component is $(2)(-1) - (-2)(1) = -2 + 2 = 0$.

Thus, our normal vector is $\vec{n} = -3\hat{i} - 3\hat{j} + 0\hat{k}$.

### Simplifying and Building the Plane

We have $\vec{n} = (-3, -3, 0)$. As we discussed, we can scale this vector to make our lives easier. Dividing by $-3$, we get a much friendlier normal vector: $\vec{n} = (1, 1, 0)$.

Now, we use the point-normal form of a plane: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$. Given the point $(1, -2, 1)$, we substitute:

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

This simplifies beautifully to $x - 1 + y + 2 = 0$, or $x + y + 1 = 0$. This is the equation of our required plane.

### The Final Distance

The last step is to find the distance from the point $(1, 2, 2)$ to our plane $x + y + 1 = 0$. We use the standard perpendicular distance formula:

$$
d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
$$

Substituting our values, we get:

$$
d = \frac{|1(1) + 1(2) + 0(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}}
$$

The numerator becomes $|1 + 2 + 0 + 1| = 4$, and the denominator is $\sqrt{2}$. So, $d = \frac{4}{\sqrt{2}}$.

Rationalizing this, we multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ to get $d = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

The final distance is **$2\sqrt{2}$**. Through the power of vector algebra, we have navigated 3D space to find the exact distance.

Animated Solution for Mathematics - Three Dimensional Geometry: A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

Select Answer:

Visualized Solution

Identify Given Planes and Normals

Condition for Perpendicularity

Setting up the Cross Product

Expanding the Determinant (i-component)

Expanding the Determinant (j and k components)

Simplifying the Normal Vector

Equation of the Plane (Point-Normal Form)

Substituting Values into Plane Equation

Finalizing the Plane Equation

The Target Point and Distance

Perpendicular Distance Formula

Substituting into Distance Formula

Calculating the Distance

Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Extracting the Normals

The Magic of the Cross Product

Simplifying and Building the Plane

The Final Distance

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