Let P1 : r · (2i + j - 3k) = 4 be a plane. Let P2 be another plane which passes through the points (

Q: Wait, why do we use the cross product of the normals? Can't we just solve the system of equations for the line?

That is a fantastic question! You absolutely *could* solve the system of equations to find the parametric form of the line. However, in JEE Advanced, time is your most precious currency. Solving the system requires finding a point on the line and then the direction. By using the cross product of the normals, we bypass the need to find a specific point on the line entirely. We go straight to the direction ratios, which is exactly what the question asks for. It is the 'surgical' approach versus the 'sledgehammer' approach!

Q: Does the order of the cross product matter? What if I did $\vec{n}_2 \times \vec{n}_1$ instead?

If you calculate $\vec{n}_2 \times \vec{n}_1$, you will get the exact negative of the correct vector. However, direction ratios are defined up to a scalar multiple. If your result was $-16, -\alpha, -\beta$, you could simply multiply by $-1$ to get $16, \alpha, \beta$. Since the problem fixes the first component at $16$, you would immediately see that your result needs to be scaled by $-1$ to match the given format. So, the physics remains consistent regardless of the order!

Q: What if the three points given for $P_2$ were collinear? Would the method still work?

If the three points were collinear, they would not define a unique plane, and the cross product $\vec{AB} \times \vec{AC}$ would result in the zero vector $\vec{0}$. In such a case, the problem would be ill-posed. However, in JEE problems, you can trust that the points provided will form a valid triangle, ensuring a unique normal vector exists. Always trust the geometry, but keep an eye out for degenerate cases!

Question

Let $P_1 : \vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 4$ be a plane. Let $P_2$ be another plane which passes through the points $(2, -3, 2), (2, -2, -3)$ and $(1, -4, 2)$. If the direction ratios of the line of intersection of $P_1$ and $P_2$ be $16, \alpha, \beta$, then the value of $\alpha + \beta$ is equal to ____.

Sigma Percentile · Accepted Answer

### The Geometry of Intersections: A 3D Odyssey

Welcome, future engineer. Today, we are not just solving a problem; we are stepping into the realm of 3D geometry. Imagine you are standing in a room where two walls meet at a corner; that corner is a line, representing the intersection of two planes.

In the JEE Advanced examination, the ability to visualize this intersection is the difference between a frantic calculation and a moment of pure, elegant clarity. Let us break down this problem as a logical journey.

### Phase 1: The Anatomy of the Planes

We are given two planes, $P_1$ and $P_2$. Our objective is to find the direction ratios of the line where they meet.

For $P_1$, the equation is given as $\vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 4$. This is the standard vector form $\vec{r} \cdot \vec{n} = d$, where $\vec{n}$ is the normal vector.

By inspection, we extract our first normal vector:
$$\vec{n}_1 = 2\hat{i} + \hat{j} - 3\hat{k}$$

Now, consider $P_2$, defined by three points: $A(2, -3, 2)$, $B(2, -2, -3)$, and $C(1, -4, 2)$. To find the normal to this plane, we construct two vectors lying within it, $\vec{AB}$ and $\vec{AC}$, and compute their cross product.

Calculating these vectors:
$$\vec{AB} = \vec{B} - \vec{A} = (2-2)\hat{i} + (-2 - (-3))\hat{j} + (-3-2)\hat{k} = 0\hat{i} + 1\hat{j} - 5\hat{k}$$
$$\vec{AC} = \vec{C} - \vec{A} = (1-2)\hat{i} + (-4 - (-3))\hat{j} + (2-2)\hat{k} = -1\hat{i} - 1\hat{j} + 0\hat{k}$$

### Phase 2: The Power of the Cross Product

To find the normal vector $\vec{n}_2$, we perform the cross product $\vec{n}_2 = \vec{AB} 	imes \vec{AC}$:

$$ \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -5 \ -1 & -1 & 0 \end{vmatrix} $$

Expanding this determinant:
For $\hat{i}$: $(1)(0) - (-5)(-1) = -5$
For $\hat{j}$: $-((0)(0) - (-5)(-1)) = 5$
For $\hat{k}$: $(0)(-1) - (1)(-1) = 1$

Thus, our second normal vector is $\vec{n}_2 = -5\hat{i} + 5\hat{j} + 1\hat{k}$.

### Phase 3: The Intersection Line

The line of intersection lies on both planes and is therefore perpendicular to both $\vec{n}_1$ and $\vec{n}_2$. The direction vector $\vec{L}$ of our line is the cross product of these two normals:

$$ \vec{L} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -3 \ -5 & 5 & 1 \end{vmatrix} $$

Computing the components:
For $\hat{i}$: $(1)(1) - (-3)(5) = 1 + 15 = 16$
For $\hat{j}$: $-((2)(1) - (-3)(-5)) = -(2 - 15) = 13$
For $\hat{k}$: $(2)(5) - (1)(-5) = 10 + 5 = 15$

The resulting direction vector is $\vec{L} = 16\hat{i} + 13\hat{j} + 15\hat{k}$.

### The Final Revelation

The problem states that the direction ratios are $16, \alpha, \beta$. By comparing our result $16, 13, 15$ with the given form, we identify $\alpha = 13$ and $\beta = 15$.

The final calculation is:
$$\alpha + \beta = 13 + 15 = \mathbf{28}$$

You have successfully navigated from abstract plane equations to specific points, utilizing the cross product to peel back the layers of this 3D geometry problem. Keep this intuition sharp, and no problem will ever be too daunting.

Enter Numerical Value:

Visualized Solution

Visualizing the Intersection of Planes

The Logic of the Intersection Line

Extracting Normal Vector $n_{1}$

Analyzing Plane $P_{2}$

Constructing Vector $A B$

Constructing Vector $A C$

Setting up Normal Vector $n_{2}$

Computing Normal Vector $n_{2}$

Setting up Direction Vector $L$

Computing Direction Vector $L$

Comparing Direction Ratios

Final Calculation

The Sigma Insight: Equation of a Plane

The Geometry of Intersections

Phase 1

Phase 2

Phase 3

The Final Revelation

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