Let P1 : 2x + y - z = 3 and P2 : x + 2y + z = 2 be two planes. Then, which of the following statemen

Q: Wait, why is the intersection line perpendicular to both normals? Shouldn't it be parallel to them?

That is a fantastic conceptual question! Think about it this way: the normal vector $\vec{n}_1$ is perpendicular to every single line lying on plane $P_1$. Since the intersection line lies on $P_1$, it must be perpendicular to $\vec{n}_1$. Similarly, it lies on $P_2$, so it must be perpendicular to $\vec{n}_2$. Therefore, the line is perpendicular to both normals simultaneously. The cross product $\vec{n}_1 \times \vec{n}_2$ is the mathematical tool that finds exactly that unique direction!

Q: Why can't I just read the direction ratios $(9, 9, 3)$ from the line equation in Statement B?

That is the classic JEE trap! The standard form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. Notice that the coefficients of $x, y,$ and $z$ must be $1$. In the given equation $\frac{3x-4}{9} = \frac{1-3y}{9} = \frac{z}{3}$, the coefficients are $3, -3,$ and $1$. You must divide the numerator and denominator of the first term by $3$ and the second term by $-3$ to reveal the true direction ratios. Always check the coefficients before jumping to conclusions!

Q: Do I really need the absolute value in the angle formula $\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$?

Yes, you absolutely do! The dot product can be negative, which would give you the cosine of an obtuse angle. However, the question asks for the 'acute angle' between the planes. The absolute value ensures that $\cos \theta$ is positive, forcing $\theta$ to be in the range $[0, 90^{\circ}]$, which is exactly what we need for an acute angle.

Question

Let $P_1 : 2x + y - z = 3$ and $P_2 : x + 2y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

Options:

(A) The line of intersection of $P_1$ and $P_2$ has direction ratios 1, 2, -1

(B) The line $\frac{3x-4}{9} = \frac{1-3y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_1$ and $P_2$

(C) The acute angle between $P_1$ and $P_2$ is $60^{\circ}$

(D) If $P_3$ is the plane passing through the point (4, 2, -2) and perpendicular to the line of intersection of $P_1$ and $P_2$, then the distance of the point (2, 1, 1) from the plane $P_3$ is $\frac{2}{\sqrt{3}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a room where two massive, flat sheets of glass intersect. We are given two planes:
$P_1: 2x + y - z = 3$
$P_2: x + 2y + z = 2$

The first step to mastering 3D geometry is to respect the normal vector. The normal vector $\vec{n}$ is the 'DNA' of a plane—it tells you exactly which way the plane is facing.

From our equations, we extract:
$\vec{n}_1 = 2\hat{i} + \hat{j} - \hat{k}$
$\vec{n}_2 = \hat{i} + 2\hat{j} + \hat{k}$

### The Spine of the Intersection

When two planes meet, they create a line. This line is trapped on both planes, meaning it must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$.

To find the direction of this spine, we deploy the cross product: $\vec{d} = \vec{n}_1 	imes \vec{n}_2$. We set up the determinant:

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

Calculating this, we get $\vec{d} = 3\hat{i} - 3\hat{j} + 3\hat{k}$. Simplifying this, we obtain the direction ratios $(1, -1, 1)$.

Looking at Statement A, which claims the ratios are $(1, 2, -1)$, we can confidently say: **Statement A is false.**

### The Trap of Standard Forms

Now, let's look at Statement B, which provides the line:
$$
\frac{3x-4}{9} = \frac{1-3y}{9} = \frac{z}{3}
$$

Many students fall into the trap of reading the denominators directly. We must rewrite this in standard form where the coefficients of $x, y,$ and $z$ are $1$:

$$
\frac{x - 4/3}{3} = \frac{y - 1/3}{-3} = \frac{z}{3}
$$

The direction vector here is $(3, -3, 3)$, which is parallel to our intersection line. Since they are parallel, they cannot be perpendicular. **Statement B is false.**

### The Angle of Intersection

Statement C asks for the acute angle between the planes. The angle between two planes is identical to the angle between their normal vectors.

We use the dot product formula:
$$
\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}
$$

Substituting our values:
$\vec{n}_1 \cdot \vec{n}_2 = (2)(1) + (1)(2) + (-1)(1) = 3$
$|\vec{n}_1| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}$
$|\vec{n}_2| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$

Thus, $\cos 	heta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$. Since $\cos 	heta = 1/2$, the angle $	heta$ is $60^{\circ}$. **Statement C is true.**

### The Final Construction

Statement D asks us to build a new plane, $P_3$, passing through $(4, 2, -2)$ and perpendicular to our intersection line. If $P_3$ is perpendicular to the line, then the line's direction vector $\vec{d} = (1, -1, 1)$ becomes the normal vector for $P_3$.

Using the point-normal form $1(x-4) - 1(y-2) + 1(z+2) = 0$, we simplify to $x - y + z = 0$. Now, we calculate the distance from $(2, 1, 1)$ to this plane using the distance formula:

$$
D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}
$$

Plugging in the numbers:
$$
D = \frac{|1(2) - 1(1) + 1(1)|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{2}{\sqrt{3}}
$$

**Statement D is true.** We have navigated the geometry, avoided the algebraic traps, and arrived at the truth.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $P_{1} : 2 x + y - z = 3$ and $P_{2} : x + 2 y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

Select Answer:

Visualized Solution

Introduction to the Planes

Direction of Intersection Line

Calculating Direction Vector $d$

Verifying Statement A

Analyzing the Line in Statement B

Checking Perpendicularity

Angle Between Planes Formula

Calculating $cos θ$

Verifying Statement C

Defining Plane $P_{3}$

Equation of Plane $P_{3}$

Distance Formula Setup

Calculating Final Distance

The Sigma Insight: Angle Between Two Planes

Analyzing the Setup

The Spine of the Intersection

The Trap of Standard Forms

The Angle of Intersection

The Final Construction

Similar Questions

Let the points on the plane $P$ be equidistant from the points $(- 4, 2, 1)$ and $(2, - 2, 3)$ . Then the acute angle between the plane $P$ and the plane $2 x + y + 3 z = 1$ is

The acute angle between the planes $P_{1}$ and $P_{2}$ , when $P_{1}$ and $P_{2}$ are the planes passing through the intersection of the planes $5 x + 8 y + 13 z - 29 = 0$ and $8 x - 7 y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$ , respectively, is

Let the acute angle bisector of the two planes $x - 2 y - 2 z + 1 = 0$ and $2 x - 3 y - 6 z + 1 = 0$ be the plane $P$ . Then which of the following points lies on $P$ ?

A plane which bisects the angle between the two given planes $2 x - y + 2 z - 4 = 0$ and $x + 2 y + 2 z - 2 = 0$ , passes through the point:

The d.r. of normal to the plane through $(1, 0, 0), (0, 1, 0)$ which makes an angle $π /4$ with plane $x + y = 3$ are

A tetrahedron has vertices $P (1, 2, 1)$ , $Q (2, 1, 3)$ , $R (- 1, 1, 2)$ and $O (0, 0, 0)$ . The angle between the faces $O P Q$ and $P QR$ is :

A tetrahedron has vertices at $O (0, 0, 0), A (1, 2, 1), B (2, 1, 3)$ and $C (- 1, 1, 2)$ . Then the angle between the faces OAB and ABC will be