The plane passing through the line L: ℓx-y+3(1-ℓ)z = 1, x+2y-z = 2 and perpendicular to the plane 3x

Q: Why do we use the 'Family of Planes' method instead of finding two points on the line?

That is a fantastic question! While finding two points on the line is theoretically possible, it is often algebraically messy and prone to error, especially when parameters like $\ell$ are involved. The 'Family of Planes' method is a more elegant, high-level approach. It allows us to treat the entire set of planes passing through the line as a single algebraic entity, which simplifies the problem significantly by letting us compare coefficients directly.

Q: Why can we equate the ratios of the coefficients of the planes?

Remember that the equation of a plane $Ax + By + Cz + D = 0$ defines its orientation in space via its normal vector $\vec{n} = (A, B, C)$. If two equations represent the same plane, their normal vectors must be parallel. Parallel vectors have proportional components. Therefore, the ratios of their coefficients must be equal. This is a fundamental property of linear equations in 3D geometry.

Q: Why do we use the cross product to find the direction of the line?

Think about the geometry: the line $L$ lies on both Plane 1 and Plane 2. This means the line is perpendicular to the normal vector of Plane 1 ($\vec{n}_1$) and also perpendicular to the normal vector of Plane 2 ($\vec{n}_2$). The only vector that is simultaneously perpendicular to two given vectors is their cross product. Thus, $\vec{d} = \vec{n}_1 \times \vec{n}_2$ gives us the direction of the line.

Question

The plane passing through the line $L: \ell x-y+3(1-\ell)z = 1, x+2y-z = 2$ and perpendicular to the plane $3x+2y+z = 6$ is $3x-8y+7z=4$. If $	heta$ is the acute angle between the line $L$ and the y-axis, then $415 \cos^2 	heta$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving an equation; we are stepping into the elegant world of 3D geometry.

Imagine standing in a room where two walls meet. That vertical edge where the walls intersect is a line. In our problem, we have two planes, $P_1$ and $P_2$, and their intersection is our line $L$.

The beauty of this problem lies in how we manipulate these planes without ever needing to 'see' the line explicitly until the very end.

### The Power of the Family of Planes

When you face a problem involving the intersection of two planes, your first instinct might be to find the coordinates of the line. Resist that urge! It is a trap that leads to tedious calculations.

Instead, we invoke the 'Family of Planes' theorem. Any plane passing through the intersection of two planes $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ can be written as $P_1 + \lambda P_2 = 0$.

By substituting our given equations, we create a single, unified equation that represents every possible plane passing through our line $L$. We are essentially creating a 'super-equation' that contains the secret of our line within its coefficients.

### The Algebraic Dance of Proportionality

Now, we are given a specific plane: $3x - 8y + 7z = 4$. We know our family of planes must contain this specific plane.

This means that for some value of $\lambda$, our general equation must be identical to the given one. In the world of linear algebra, two equations represent the same plane if and only if their coefficients are proportional.

This is where we set up our ratios:

$$ \frac{\ell + \lambda}{3} = \frac{2\lambda - 1}{-8} = \frac{3(1-\ell) - \lambda}{7} = \frac{-(1 + 2\lambda)}{-4} $$

This looks intimidating, but take a deep breath. It is just a system of linear equations. By isolating the ratios, we can solve for $\lambda$ and $\ell$ with precision.

We find $\lambda = -\frac{1}{6}$ and $\ell = \frac{2}{3}$. The algebra is just the vehicle; the geometry is the destination.

### Finding the Direction of the Line

With $\ell$ determined, we have fully defined our two planes. Now, we need the direction of the line $L$.

Recall that the line $L$ is the intersection of these two planes. This means the line must be perpendicular to the normal vector of the first plane, $\vec{n}_1$, and the normal vector of the second plane, $\vec{n}_2$.

How do we find a vector perpendicular to two others? The cross product! We compute $\vec{d} = \vec{n}_1 	imes \vec{n}_2$.

Setting up the determinant with $\hat{i}, \hat{j}, \hat{k}$ and expanding it gives us the direction vector. After simplifying, we get:

$$ \vec{d} = -\hat{i} + \frac{5}{3}\hat{j} + \frac{7}{3}\hat{k} $$

To make our lives easier, we multiply by 3 to get the cleaner direction ratios $(-3, 5, 7)$.

### The Final Angle

We are almost there. We need the acute angle $	heta$ between our line $L$ and the y-axis. The y-axis has the direction vector $\vec{j} = (0, 1, 0)$.

The cosine of the angle between two vectors is given by the dot product formula:

$$ \cos 	heta = \frac{|\vec{d} \cdot \vec{j}|}{|\vec{d}| |\vec{j}|} $$

Substituting our values, we get $\cos 	heta = \frac{5}{\sqrt{83}}$. The question asks for $415 \cos^2 	heta$.

Squaring our result gives $\frac{25}{83}$. Multiplying by 415, the 83 in the denominator cancels out perfectly with the 415, leaving us with $5 	imes 25 = \mathbf{125}$.

### Conclusion

Look at what we have achieved. We navigated through the family of planes, mastered the proportionality of coefficients, utilized the cross product to find direction, and finished with a clean dot product.

This is the essence of JEE Advanced physics and math—not just calculation, but the orchestration of concepts. You have mastered the geometry of the line. Keep this confidence, and carry it into your next challenge.

Enter Numerical Value:

Visualized Solution

Intersection of Planes

Family of Planes

Grouping Coefficients

Comparing Planes

Proportional Ratios

Solving for $λ$

Solving for $ℓ$

Direction of Line $L$

Cross Product

Simplifying Direction Ratios

Angle with Y-axis

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Power of the Family of Planes

The Algebraic Dance of Proportionality

Finding the Direction of the Line

The Final Angle

Conclusion

Similar Questions

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