The acute angle between the planes P1 and P2, when P1 and P2 are the planes passing through the inte

Q: Wait, why do we use two different parameters, $\lambda$ and $\mu$, for the two planes? Can't we just use one?

That is a fantastic question! Think of the family of planes as an infinite set of possibilities. Each unique value of the parameter selects exactly one plane from that infinite set. Since $P_1$ and $P_2$ are distinct planes, they correspond to two distinct values of the parameter. Using $\lambda$ and $\mu$ helps us keep track of these two specific, separate identities.

Q: Why do we calculate the angle between the normal vectors instead of the planes themselves?

Geometry is all about finding the easiest path to the truth. Measuring the angle between two flat surfaces directly is difficult. However, the normal vector is a unique 'fingerprint' of the plane's orientation. If you tilt the plane, the normal tilts by the exact same amount. Thus, the angle between the normals is geometrically equivalent to the angle between the planes.

Q: Why do we take the absolute value in the dot product formula? What if the result is negative?

The dot product can indeed be negative, which would give us the obtuse angle between the normal vectors. But the question specifically asks for the *acute* angle between the planes. By taking the absolute value, we ensure our cosine value is positive, which forces our result to be in the first quadrant—giving us exactly the acute angle we need.

Question

The acute angle between the planes $P_1$ and $P_2$, when $P_1$ and $P_2$ are the planes passing through the intersection of the planes $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$, respectively, is

Options:

(A) $\frac{\pi}{3}$

(B) $\frac{\pi}{4}$

(C) $\frac{\pi}{6}$

(D) $\frac{\pi}{12}$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Infinite Family

Imagine you are standing in a vast, empty room with two flat, infinite sheets of paper representing planes $L_1$ and $L_2$. Because they are not parallel, they must intersect to form a single, perfectly straight line—the spine of a book.

An infinite number of pages bound to that spine creates a 'family of planes'. In JEE Advanced geometry, we represent every plane in this family using the elegant equation:

$$L_1 + \lambda L_2 = 0$$

Here, $\lambda$ is a parameter. By varying $\lambda$, we rotate through the entire family of planes passing through the common line of intersection.

### Pinning Down Our Targets

We seek two specific planes, $P_1$ and $P_2$, from this family. We are given that $P_1$ passes through $(2, 1, 3)$ and $P_2$ passes through $(0, 1, 2)$.

Since these points lie on their respective planes, they must satisfy the family equation:

$$(5x + 8y + 13z - 29) + \lambda(8x - 7y + z - 20) = 0$$

For $P_1$, substituting $(2, 1, 3)$ yields:

$$(10 + 8 + 39 - 29) + \lambda(16 - 7 + 3 - 20) = 0$$

Solving this gives $\lambda = \frac{7}{2}$. Substituting this back and simplifying, we arrive at the equation for $P_1$:

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

We repeat this process for $P_2$ using a parameter $\mu$. Substituting $(0, 1, 2)$ into the family equation:

$$(0 + 8 + 26 - 29) + \mu(0 - 7 + 2 - 20) = 0$$

This yields $\mu = \frac{1}{5}$. Simplifying the resulting expression gives the equation for $P_2$:

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

### The Bridge of Normal Vectors

To find the angle between these two planes, we examine their normal vectors. A plane defined by $ax + by + cz + d = 0$ has a normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$.

For $P_1$, the normal is $\vec{n}_1 = 2\hat{i} - \hat{j} + \hat{k}$. For $P_2$, the normal is $\vec{n}_2 = \hat{i} + \hat{j} + 2\hat{k}$.

The angle $	heta$ between the planes is equivalent to the angle between these normal vectors. We use the dot product formula:

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

### The Final Calculation

First, we calculate the dot product:

$$\vec{n}_1 \cdot \vec{n}_2 = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$$

Next, we find the magnitudes:

$$|\vec{n}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$$
$$|\vec{n}_2| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$$

Substituting these into the cosine formula:

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

Taking the acute angle, we find $	heta = \cos^{-1}(\frac{1}{2})$, which results in:

$$	heta = \frac{\pi}{3}$$

This result demonstrates how complex 3D geometry can be reduced to elegant, manageable components through the use of family equations and normal vectors.

Select Answer:

Visualized Solution

Visualizing the Family of Planes

Equation of the Family of Planes

Targeting Plane $P_{1}$

Solving for $λ$

Final Equation of $P_{1}$

Targeting Plane $P_{2}$

Solving for $μ$

Final Equation of $P_{2}$

Extracting the Normal Vectors

Formula for Angle Between Planes

Calculating Dot Product and Magnitudes

Evaluating the Values

Final Angle Calculation

The Sigma Insight: Angle Between Two Planes

Analyzing the Geometry of the Infinite Family

Pinning Down Our Targets

The Bridge of Normal Vectors

The Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Family of Planes

Equation of the Family of Planes

Targeting Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​

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

Final Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​

Targeting Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​

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

Final Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​