Let the plane P : r · a = d contain the line of intersection of two planes r · (i + 3j - k) = 6 and

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

That is a fantastic question! While you *could* find the line of intersection, it is a much longer, more tedious process involving finding the direction vector of the line (cross product of the normals) and a point on the line. The 'family of planes' method is a powerful shortcut in JEE Advanced. It directly gives you the equation of the required plane without needing to explicitly define the line. Efficiency is key in competitive exams!

Q: Why do we convert the vector equations to Cartesian form? Can't we just work with the vector form?

You absolutely can work with the vector form! However, when you are given a specific point like $(2, 3, \frac{1}{2})$, Cartesian coordinates are much easier to substitute into the equation. It reduces the risk of algebraic errors. In the end, both methods lead to the same result, but Cartesian form is often more intuitive for point substitution.

Q: How do I know if the constant $d$ in $\vec{r} \cdot \vec{a} = d$ is the same as the constant term in my Cartesian equation?

Great observation. In the standard form $\vec{r} \cdot \vec{a} = d$, the constant $d$ is on the right-hand side. If your Cartesian equation is $Ax + By + Cz = D$, then $D$ is your $d$. Just ensure your equation is in the form $Ax + By + Cz = D$ before comparing it to the vector form. If you have $Ax + By + Cz - D = 0$, just move the constant to the other side!

Question

Let the plane $P : \vec{r} \cdot \vec{a} = d$ contain the line of intersection of two planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$ and $\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$. If the plane $P$ passes through the point $(2, 3, \frac{1}{2})$, then the value of $\frac{|13\vec{a}|^2}{d^2}$ is equal to

Options:

(A) 90

(B) 93

(C) 95

(D) 97

Sigma Percentile · Accepted Answer

### The Geometry of the Hinge

Imagine you are standing in a 3D space, and you see two massive, infinite planes intersecting. Where they meet, they form a perfectly straight line—a hinge.

Now, imagine you want to construct a third plane that passes exactly through this hinge. There are infinitely many such planes, all rotating around that single line of intersection.

This is what we call a 'family of planes'. In JEE Advanced geometry, mastering this concept is like having a skeleton key for 3D coordinate geometry problems.

### The Algebraic Tool: The Family of Planes

We are given two planes in vector form:

$$P_1: \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$$

$$P_2: \vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$$

To make our lives easier, let's translate these into the language of Cartesian coordinates. By substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, we get:

$$P_1: x + 3y - z - 6 = 0$$

$$P_2: -6x + 5y - z - 7 = 0$$

Now, the magic happens. The equation of any plane passing through the intersection of $P_1$ and $P_2$ can be written as:

$$(x + 3y - z - 6) + \lambda(-6x + 5y - z - 7) = 0$$

Here, $\lambda$ is our scalar parameter. It acts like a dial, allowing us to select any specific plane from the infinite family of planes that share that hinge.

### Finding the Specificity

We aren't looking for just *any* plane; we are looking for the one that passes through the point $(2, 3, \frac{1}{2})$. This is our anchor.

By substituting these coordinates into our family equation, we can lock in the value of $\lambda$:

$$(2 + 3(3) - \frac{1}{2} - 6) + \lambda(-6(2) + 5(3) - \frac{1}{2} - 7) = 0$$

Let's take a breath and calculate this carefully. The first bracket simplifies to $11 - 6.5 = 4.5$.

The second bracket simplifies to $-12 + 15 - 0.5 - 7 = 3 - 7.5 = -4.5$. So, we have:

$$4.5 + \lambda(-4.5) = 0$$

This leads us to the elegant result: $\lambda = 1$.

### The Final Elegance

With $\lambda = 1$, our plane equation becomes simple addition:

$$(x + 3y - z - 6) + 1(-6x + 5y - z - 7) = 0$$

$$-5x + 8y - 2z - 13 = 0$$

Rearranging this to match the standard form $\vec{r} \cdot \vec{a} = d$, we get $-5x + 8y - 2z = 13$. Thus, our normal vector is $\vec{a} = -5\hat{i} + 8\hat{j} - 2\hat{k}$ and our constant is $d = 13$.

Finally, we evaluate the expression $\frac{|13\vec{a}|^2}{d^2}$. Since $d = 13$, the expression simplifies beautifully to $|\vec{a}|^2$.

We calculate the squared magnitude of $\vec{a}$:

$$|\vec{a}|^2 = (-5)^2 + 8^2 + (-2)^2 = 25 + 64 + 4 = 93$$

And there we have it! The complexity of the intersection, the family of planes, and the point substitution all collapse into the clean, satisfying integer **93**. Keep practicing these visualizations—they are the foundation of your success in JEE Advanced.

Select Answer:

Visualized Solution

Visualizing the Intersection

Converting to Cartesian Form

The Family of Planes Concept

Using the Given Point

Solving for $λ$

Finding the Equation of Plane $P$

Identifying $a$ and $d$

The Final Expression Setup

Final Calculation

The Sigma Insight: Equation of a Plane

The Geometry of the Hinge

The Algebraic Tool

Finding the Specificity

The Final Elegance

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