The vector equation of the plane passing through the intersection of the planes r·(i+j+k) = 1 and r·

Q: Why do we use the family of planes equation $P_1 + \lambda P_2 = 0$ instead of finding the line of intersection first?

Finding the line of intersection is a valid path, but it is computationally heavy and prone to errors. The family of planes method is an elegant shortcut. It treats the entire line as a single entity, allowing us to bypass the tedious work of finding the line's direction vector and a point on it.

Q: What if the point $(1, 0, 2)$ didn't satisfy the equation? Does that mean the point is not on the plane?

Exactly! If a point does not satisfy the equation, it simply does not lie on that plane. In this problem, the point is a constraint given to us. By forcing the plane to pass through this point, we are effectively 'locking' the value of $\lambda$ to select the one specific plane that contains that point.

Q: Is the order of $P_1$ and $P_2$ important in the equation $P_1 + \lambda P_2 = 0$?

Not at all. You could write it as $\lambda P_1 + P_2 = 0$ and you would arrive at the same final plane. The only difference is that your $\lambda$ value would be the reciprocal of the one we found. The final equation of the plane will remain identical.

Question

The vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} - 2\hat{j}) = -2$, and the point $(1,0,2)$ is :

Options:

(A) $\vec{r} \cdot (\hat{i} - 7\hat{j} + 3\hat{k}) = \frac{7}{3}$

(B) $\vec{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$

(C) $\vec{r} \cdot (3\hat{i} + 7\hat{j} + 3\hat{k}) = 7$

(D) $\vec{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = \frac{7}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of the Book Spine: Mastering the Family of Planes

Welcome, future engineer. Today, we are not just solving a problem; we are visualizing the architecture of three-dimensional space. Imagine two planes intersecting in space.

Just like the pages of an open book meeting at the spine, these two planes create a unique line of intersection. Our mission is to find the equation of a new plane that passes exactly through this spine. This is a classic JEE Advanced problem that tests your ability to handle the **Family of Planes** concept.

### The Concept of the Family

When we have two planes, $P_1 = 0$ and $P_2 = 0$, any plane passing through their intersection can be represented by the equation $P_1 + \lambda P_2 = 0$. Think of $\lambda$ as a tuning knob.

As you change the value of $\lambda$, you are essentially rotating a plane around the spine of the book. Every value of $\lambda$ gives you a different page in this infinite book. Our goal is to find the specific $\lambda$ that gives us the page passing through the point $(1, 0, 2)$.

### Setting the Stage

We are given the planes $P_1: \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 = 0$ and $P_2: \vec{r} \cdot (\hat{i} - 2\hat{j}) + 2 = 0$. By substituting these into our family equation, we get:

$$(\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1) + \lambda (\vec{r} \cdot (\hat{i} - 2\hat{j}) + 2) = 0$$

This looks intimidating, but it is just a linear equation waiting for a value.

### The Constraint

We are told the plane passes through the point $(1, 0, 2)$. In vector notation, the position vector of this point is $\vec{a} = \hat{i} + 2\hat{k}$. Since this point lies on our plane, it must satisfy the equation.

We substitute $\vec{r} = \hat{i} + 2\hat{k}$ into our family equation. Let's evaluate the dot products carefully. For the first part:

$$(\hat{i} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 = (1 + 0 + 2) - 1 = 2$$

For the second part:

$$(\hat{i} + 2\hat{k}) \cdot (\hat{i} - 2\hat{j}) + 2 = (1 + 0 + 0) + 2 = 3$$

Now, our equation simplifies to $2 + 3\lambda = 0$. Solving this, we find $\lambda = -\frac{2}{3}$.

### The Final Assembly

Now that we have our $\lambda$, we substitute it back into the family equation:

$$(\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1) - \frac{2}{3} (\vec{r} \cdot (\hat{i} - 2\hat{j}) + 2) = 0$$

To make this elegant, we multiply the entire equation by $3$ to clear the fraction:

$$3(\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1) - 2(\vec{r} \cdot (\hat{i} - 2\hat{j}) + 2) = 0$$

Expanding this, we get:

$$\vec{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) - 3 - \vec{r} \cdot (2\hat{i} - 4\hat{j}) - 4 = 0$$

Grouping the terms, we arrive at the final result:

$$\vec{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$$

This is the equation of our plane. It is beautiful, precise, and perfectly aligned with the geometry of the problem. Remember, in JEE Advanced, the math is just the language; the visualization is the story.

Animated Solution for Mathematics - Three Dimensional Geometry: The vector equation of the plane passing through the intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (\hat{i} - 2 \hat{j}) = - 2$ , and the point $(1, 0, 2)$ is :

Select Answer:

Visualized Solution

Intersection of Two Planes

Family of Planes

Setting up the Equation

The Constraint Point

Substituting the Point

Evaluating $P_{1}$ at the Point

Evaluating $P_{2}$ at the Point

Solving for $λ$

Substituting $λ$ back

Expanding the Terms

Final Vector Equation

The Sigma Insight: Equation of a Plane

The Geometry of the Book Spine

The Concept of the Family

Setting the Stage

The Constraint

The Final Assembly

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