Let the plane passing through the point (-1, 0, -2) and perpendicular to each of the planes 2x + y -

Q: Why do we use the cross product? Can't we just use dot products?

Great question! The dot product of two vectors gives a scalar, which tells us about their angle, but it doesn't give us a new vector. The cross product, however, is specifically designed to produce a vector that is mutually perpendicular to the two input vectors. Since our target plane is perpendicular to both given planes, its normal must be perpendicular to both of their normals. The cross product is the perfect tool for this!

Q: Why did we multiply the final equation by $-1$?

The problem asks for the equation in the form $ax + by + cz + 8 = 0$. After our initial calculation, we arrived at $-2x + y - 3z - 8 = 0$. To match the constant term $+8$ required by the problem statement, we simply multiply the entire equation by $-1$, which gives us $2x - y + 3z + 8 = 0$. This doesn't change the plane itself, just its representation.

Q: Is the normal vector unique? What if I got a different one?

The normal vector is not unique! Any scalar multiple of the normal vector represents the same plane. For example, if you had calculated the normal as $(2, -1, 3)$ or $(-4, 2, -6)$, they would all define the same plane. However, the ratio of the coefficients $a:b:c$ will always be the same. As long as your final equation is a multiple of the correct one, you are on the right track.

Question

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

Options:

(A) 3

(B) 8

(C) 5

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the 3D realm! Today, we are tackling a classic JEE Advanced problem that tests not just your algebraic skills, but your ability to visualize the architecture of space.

We are tasked with finding the equation of a plane that passes through a specific point $P(-1, 0, -2)$ and stands perpendicular to two other planes. It sounds daunting, but let's break it down into a beautiful, logical journey.

### The DNA of a Plane

Every plane in 3D space has a unique identifier: its normal vector. Think of the normal vector as the plane's DNA, as it tells you exactly how the plane is oriented.

For Plane 1, given by $2x + y - z = 2$, the coefficients of $x, y,$ and $z$ give us its normal vector, $\vec{n}_1 = (2, 1, -1)$.

Similarly, for Plane 2, $x - y - z = 3$, the normal vector is $\vec{n}_2 = (1, -1, -1)$. These vectors are the keys to our solution.

### The Perpendicularity Engine

Here is the core geometric insight: if our target plane is perpendicular to Plane 1 and Plane 2, then its normal vector $\vec{n}$ must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$.

In the language of vectors, we need a vector that is orthogonal to two given vectors. This is the exact definition of the cross product, so we calculate $\vec{n} = \vec{n}_1 	imes \vec{n}_2$.

To compute this, we set up a determinant:

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

Expanding this along the first row, we get:

$$\vec{n} = \hat{i}(1(-1) - (-1)(-1)) - \hat{j}(2(-1) - 1(-1)) + \hat{k}(2(-1) - 1(1))$$

Calculating each component carefully: for $\hat{i}$, we have $-1 - 1 = -2$. For $\hat{j}$, we have $-( -2 + 1) = -(-1) = 1$. For $\hat{k}$, we have $-2 - 1 = -3$.

Thus, our normal vector is $\vec{n} = -2\hat{i} + \hat{j} - 3\hat{k}$.

### Constructing the Plane

Now that we have our normal vector $\vec{n} = (-2, 1, -3)$ and the point $P(-1, 0, -2)$, we use the point-normal form of a plane equation: $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.

Substituting our values, we get:

$$-2(x - (-1)) + 1(y - 0) - 3(z - (-2)) = 0$$

$$-2(x + 1) + y - 3(z + 2) = 0$$

$$-2x - 2 + y - 3z - 6 = 0$$

$$-2x + y - 3z - 8 = 0$$

### The Final Polish

The problem asks for the form $ax + by + cz + 8 = 0$. Our current equation has a constant of $-8$.

To fix this, we multiply the entire equation by $-1$ to get $2x - y + 3z + 8 = 0$. Comparing this to the target form, we identify $a = 2, b = -1,$ and $c = 3$.

The final step is to calculate the sum **$a + b + c = 2 - 1 + 3 = 4$**.

And there you have it! By visualizing the geometric relationship and carefully executing the vector algebra, we have unlocked the solution. Keep practicing this spatial visualization—it is the secret to mastering 3D geometry in JEE Advanced!

Animated Solution for Mathematics - Three Dimensional Geometry: 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:

Select Answer:

Visualized Solution

Visualizing the Problem

Identifying Normals $n_{1}$ and $n_{2}$

The Geometry of Perpendicularity

Setting up the Cross Product

Expanding the Determinant

Finding the Normal $n$

Point-Normal Form of a Plane

Substituting the Values

Expanding the Equation

Matching the Target Form

Identifying $a, b,$ and $c$

Calculating the Final Sum

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The DNA of a Plane

The Perpendicularity Engine

Constructing the Plane

The Final Polish

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