Let θ be the angle between the planes P1 = r ⋅ (i + j + 2k) = 9 and P2 = r ⋅ (2i - j + k) = 15. Let

Q: Wait, why did we ignore the point (4, -2, 5) entirely? Is it not important?

That is a brilliant observation! In JEE Advanced, examiners love to include 'noise'—extra data that isn't required for the specific calculation. Since the question asks for a ratio of trigonometric functions of angles, the spatial position of the line is irrelevant; only its orientation matters.

Q: Why is the angle between two planes defined by their normals? Can't we just look at the planes directly?

Think of a plane as a flat sheet of paper. It is hard to measure the angle between two sheets directly. However, if you stick a pencil (the normal vector) perpendicular to each sheet, the angle between the pencils is exactly the same as the angle between the sheets. It is a beautiful geometric shortcut!

Q: How do I know if I should use $\cos \theta$ or $\sin \theta$ for the angle between a line and a plane?

Always draw a small diagram. The normal is at $90^\circ$ to the plane. If the line makes angle $\theta$ with the normal, the remaining angle to the plane must be $90^\circ - \theta$. Thus, the angle with the plane is the complement of the angle with the normal.

Q: Does the dot product formula for the angle between planes always use the absolute value?

Yes! By convention, the angle between two planes is taken as the acute angle. Using the absolute value $|\vec{n}_1 \cdot \vec{n}_2|$ ensures that your cosine value is positive, leading to an angle between $0^\circ$ and $90^\circ$.

Q: Is there any scenario where $\alpha + \theta$ would not be $90^\circ$?

Only if the line is not defined relative to the normal. But by definition, the normal is perpendicular to every line lying in the plane. Therefore, the angle between the normal and any line in the plane is $90^\circ$. The geometry is rigid and absolute!

Question

Let $	heta$ be the angle between the planes $P_1 = \vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 9$ and $P_2 = \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 15$. Let $L$ be the line that meets $P_2$ at the point $(4, -2, 5)$ and makes an angle $	heta$ with the normal of $P_2$. If $\alpha$ is the angle between $L$ and $P_2$ then $(	an^2 	heta)(\cot^2 \alpha)$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are navigating the elegant architecture of 3D space. When you look at the equations of planes like $P_1: \vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 9$ and $P_2: \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 15$, do not see them as mere algebraic constraints.

See them as two infinite sheets slicing through the universe, each defined by a unique 'direction'—their normal vectors.

### The Soul of the Plane: The Normal Vector

Every plane has a heartbeat, and that heartbeat is its normal vector. For $P_1$, the normal is $\vec{n}_1 = \hat{i} + \hat{j} + 2\hat{k}$. For $P_2$, it is $\vec{n}_2 = 2\hat{i} - \hat{j} + \hat{k}$.

These vectors are the 'anchors' of our geometry. When we ask for the angle $	heta$ between two planes, we are essentially asking: "How much do these two anchors tilt away from each other?"

To find this, we use the dot product, the bridge between algebra and geometry:

$$\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$$

Calculating the dot product $\vec{n}_1 \cdot \vec{n}_2 = (1)(2) + (1)(-1) + (2)(1) = 3$. Finding the magnitudes, we have $|\vec{n}_1| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$ and $|\vec{n}_2| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.

Substituting these into our formula, we arrive at:

$$\cos 	heta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$$

This reveals that $	heta = 60^\circ$. The planes are tilted at a perfect $60^\circ$ angle to one another.

### The Line and the Plane: A Complementary Dance

Now, consider the line $L$. The problem states it makes an angle $	heta$ with the normal of $P_2$. Imagine the normal vector standing tall like a flagpole on the plane.

If our line $L$ leans away from that flagpole by $	heta = 60^\circ$, how much does it lean toward the ground (the plane)?

Because the flagpole is perpendicular to the ground, the relationship is strictly complementary: $\alpha + 	heta = 90^\circ$. With $	heta = 60^\circ$, we find that $\alpha = 30^\circ$.

This is the beauty of 3D geometry—everything is interconnected by simple, rigid rules.

### The Final Synthesis

We are asked to evaluate $(	an^2 	heta)(\cot^2 \alpha)$. Substituting our values:

$$	an^2 60^\circ = (\sqrt{3})^2 = 3$$

$$\cot^2 30^\circ = (\sqrt{3})^2 = 3$$

Multiplying these together, we get:

$$3 	imes 3 = 9$$

Look at that result. It is clean, it is precise, and it is the reward for your patience. You didn't need the point $(4, -2, 5)$ to reach this truth.

In JEE Advanced, the most powerful tool you possess is not just your calculator, but your ability to strip away the unnecessary and focus on the geometric essence. You have mastered the orientation of planes and lines today. The final answer is **9**.

Enter Numerical Value:

Visualized Solution

Visualize the Planes $P_{1}$ and $P_{2}$

Identify Normal Vectors $n_{1}$ and $n_{2}$

Formula for Angle $θ$

Calculate Dot Product $n_{1} \cdot n_{2}$

Calculate Magnitudes $∣ n_{1} ∣$ and $∣ n_{2} ∣$

Find $cos θ$ and $θ$

Analyze Line $L$ and Angle $α$

Relationship Between $α$ and $θ$

Calculate $α$

Setup Final Expression

Substitute and Calculate Final Answer

The Sigma Insight: Angle Between Two Planes

Analyzing the Setup

The Soul of the Plane

The Line and the Plane

The Final Synthesis

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Enter Numerical Value:

Visualized Solution

Visualize the Planes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​

Identify Normal Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​

Formula for Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Calculate Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​⋅n2​

Calculate Magnitudes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣n1​∣ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣n2​∣

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Analyze Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L and Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Relationship Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α