If the equation of the plane passing through the point (1, 1, 2) and perpendicular to the line x - 3

Q: Wait, why is the direction of the line the same as the normal vector of the plane? I thought they were different things.

That is a fantastic question! Think of it this way: if a plane is perpendicular to a line, the line is 'sticking out' of the plane at a 90-degree angle. By definition, the normal vector of a plane is any vector that is perpendicular to the plane's surface. Since the line is perpendicular to the surface, its direction vector is, by definition, a normal vector to that plane. They are essentially the same direction!

Q: Why do we have to divide the whole equation by 28 at the end? Can't we just leave it as $-x + 7y + 11z = 28$?

You absolutely could leave it that way, but the problem specifically asks for the form $Ax + By + Cz = 1$. In competitive exams like JEE, the format is part of the constraint. By dividing by 28, we are 'normalizing' the equation to fit the specific algebraic structure requested, which allows us to extract the values of $A$, $B$, and $C$ directly.

Q: Is there a faster way to find the line of intersection without using the cross product?

While you could solve the system of two equations with three variables by parameterizing one variable (e.g., setting $z = k$), the cross product method is generally faster and less prone to algebraic errors in this specific context. It directly gives you the direction ratios $(a, b, c)$ without needing to find two points on the line.

Question

If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the line $x - 3y + 2z - 1 = 0 = 4x - y + z$ is $Ax + By + Cz = 1$, then $140(C - B + A)$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are stepping into the realm of three-dimensional space. Many students fear 3D geometry because it requires us to move beyond the flat surface of a notebook and visualize objects interacting in space.

But fear not—once you grasp the underlying logic, these problems become as elegant as a well-composed symphony.

Imagine two sheets of paper intersecting in mid-air. Where they meet, they form a straight line, which we define as line $L$. We are given two planes:

$$Plane 1: x - 3y + 2z - 1 = 0$$

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

Our goal is to find a third plane. This new plane is perpendicular to the line $L$ formed by the intersection of the first two planes. We are also given a point $P(1, 1, 2)$ that lies on this new plane.

To define a plane, we need two fundamental pieces of information: a point on the plane and a normal vector $\vec{n}$. We already possess the point $P$.

### The Power of the Cross Product

The core insight is this: if our target plane is perpendicular to line $L$, then the direction of line $L$ is exactly the direction of the normal vector $\vec{n}$ of our plane.

Since line $L$ lies on both Plane 1 and Plane 2, it must be perpendicular to the normal vectors of both planes. Let $\vec{n_1}$ be the normal of Plane 1 and $\vec{n_2}$ be the normal of Plane 2. From the equations, we extract:

$$\vec{n_1} = \hat{i} - 3\hat{j} + 2\hat{k}$$

$$\vec{n_2} = 4\hat{i} - \hat{j} + \hat{k}$$

To find a vector that is perpendicular to both $\vec{n_1}$ and $\vec{n_2}$, we use the cross product. Let the direction vector of the line be $\vec{v} = \vec{n_1} 	imes \vec{n_2}$.

$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 2 \ 4 & -1 & 1 \end{vmatrix}$$

Expanding this determinant, we obtain:

$$\vec{v} = \hat{i}(-3 - (-2)) - \hat{j}(1 - 8) + \hat{k}(-1 - (-12))$$

$$\vec{v} = -\hat{i} + 7\hat{j} + 11\hat{k}$$

Thus, our normal vector is $\vec{n} = (-1, 7, 11)$. We have successfully translated the geometric relationship into a concrete vector.

### Constructing the Plane

Now that we have the normal vector $\vec{n} = (-1, 7, 11)$ and the point $P(1, 1, 2)$, we use the standard point-normal form of a plane equation:

$$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$

Substituting our values into the equation:

$$-1(x - 1) + 7(y - 1) + 11(z - 2) = 0$$

Expanding this, we get:

$$-x + 1 + 7y - 7 + 11z - 22 = 0$$

$$-x + 7y + 11z - 28 = 0$$

Rearranging the terms, we find the equation of the plane:

$$-x + 7y + 11z = 28$$

### Final Calculation

The problem asks us to match the form $Ax + By + Cz = 1$. To achieve this, we divide the entire equation by 28:

$$-\frac{1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1$$

By comparing this to $Ax + By + Cz = 1$, we identify our coefficients:

$$A = -\frac{1}{28}, \quad B = \frac{7}{28}, \quad C = \frac{11}{28}$$

Finally, we calculate the requested expression $140(C - B + A)$:

$$140 \left( \frac{11}{28} - \frac{7}{28} + \left(-\frac{1}{28}ight) ight)$$

$$140 \left( \frac{11 - 7 - 1}{28} ight) = 140 \left( \frac{3}{28} ight)$$

Since $140 \div 28 = 5$, we are left with $5 	imes 3 = 15$. The final answer is **15**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the line $x - 3 y + 2 z - 1 = 0 = 4 x - y + z$ is $A x + B y + C z = 1$ , then $140 (C - B + A)$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Visualizing the 3D Geometry Setup

Identifying the Normal Vector

Direction of the Intersection Line

Extracting Normal Vectors

Setting up the Cross Product

Calculating the Direction Vector

Setting up the Plane Equation

Expanding the Equation

Simplifying the Equation

Normalizing the Equation

Finding A, B, and C

Setting up the Final Expression

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Cross Product

Constructing the Plane

Final Calculation

Similar Questions

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

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

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A plane $E$ is perpendicular to the two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $32$ , then $(P Q)^{2}$ is equal to

Let the equation of the plane passing through the line $x - 2 y - z - 5 = 0 = x + y + 3 z - 5$ and parallel to the line $x + y + 2 z - 7 = 0 = 2 x + 3 y + z - 2$ be $a x + b y + cz = 65$ . Then the distance of the point $(a, b, c)$ from the plane $2 x + 2 y - z + 16 = 0$ is ______.

Let the equation of the plane $P$ containing the line $\frac{x + 10}{1} = \frac{8 - y}{2} = z$ be $a x + b y + 3 z = 2 (a + b)$ and the distance of the plane $P$ from the point $(1, 27, 7)$ be $c$ . Then $a^{2} + b^{2} + c^{2}$ is equal to ______.

Let $(λ, 2, 1)$ be a point on the plane which passes through the point $(4, - 2, 2)$ . If the plane is perpendicular to the line joining the point $(- 2, - 21, 29)$ and $(- 1, - 16, 23)$ , then $(\frac{λ}{11})^{2} - \frac{4 λ}{11} - 4$ is equal to

The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ , having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is:

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

Enter Numerical Value:

Visualized Solution

Visualizing the 3D Geometry Setup

Identifying the Normal Vector

Direction of the Intersection Line

Extracting Normal Vectors

Setting up the Cross Product

Calculating the Direction Vector

Setting up the Plane Equation

Expanding the Equation

Simplifying the Equation

Normalizing the Equation

Finding A, B, and C

Setting up the Final Expression

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Cross Product

Constructing the Plane

Final Calculation

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