Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection w

Q: Wait, why do we use the family of planes equation $P_1 + \lambda P_2 = 0$?

Think of the line of intersection as a hinge. Any plane that passes through this hinge can be described by this linear combination. It is a powerful tool in 3D geometry because it allows us to represent an infinite number of planes passing through a single line using just one parameter, $\lambda$.

Q: How do we know the new plane is perpendicular to the original one?

The problem explicitly states the plane is rotated through a 'right angle' (90 degrees). If a plane is rotated by 90 degrees, its orientation changes such that its normal vector becomes perpendicular to the original normal vector. This is a geometric property: the angle between the planes is the same as the angle between their normals.

Q: Is there a faster way to find the mirror image without the long formula?

While the formula $\frac{x_2 - x_1}{l} = \frac{y_2 - y_1}{m} = \frac{z_2 - z_1}{n} = \frac{-2(lx_1 + my_1 + nz_1 + d)}{l^2 + m^2 + n^2}$ looks intimidating, it is the most robust and reliable method for JEE. It ensures you account for both the distance from the plane and the direction of the normal vector correctly. Don't skip steps; precision is your best friend here.

Question

Let the plane $2x + 3y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3y + 5z = 8$. If the mirror image of the point $(2, -\frac{1}{2}, 2)$ in the rotated plane is $B(a, b, c)$, then :

Options:

(A) $\frac{a}{8} = \frac{b}{5} = \frac{c}{-4}$

(B) $\frac{a}{4} = \frac{b}{5} = \frac{c}{-2}$

(C) $\frac{a}{8} = \frac{b}{-5} = \frac{c}{4}$

(D) $\frac{a}{4} = \frac{b}{5} = \frac{c}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The intersection of two planes $P_1: 2x + 3y + z + 20 = 0$ and $P_2: x - 3y + 5z - 8 = 0$ defines a hinge line. Any plane passing through this line belongs to the family of planes defined by $P_1 + \lambda P_2 = 0$.

Substituting the given equations, we obtain:

$$(2x + 3y + z + 20) + \lambda(x - 3y + 5z - 8) = 0$$

Grouping the coefficients of $x, y,$ and $z$, the equation of the new plane $P_3$ is:

$$(2 + \lambda)x + (3 - 3\lambda)y + (1 + 5\lambda)z + (20 - 8\lambda) = 0$$

### The Perpendicularity Condition

The normal vector of $P_1$ is $\vec{n}_1 = (2, 3, 1)$. The normal vector of our new plane $P_3$ is $\vec{n}_3 = (2 + \lambda, 3 - 3\lambda, 1 + 5\lambda)$.

Since $P_3$ is perpendicular to $P_1$, their dot product must be zero: $\vec{n}_1 \cdot \vec{n}_3 = 0$.

$$2(2 + \lambda) + 3(3 - 3\lambda) + 1(1 + 5\lambda) = 0$$

Expanding the terms, we get:

$$4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambda = 0$$

Simplifying this expression yields $14 - 2\lambda = 0$, which results in $\lambda = 7$.

### The Rotated Plane

Substituting $\lambda = 7$ back into the family equation:

$$(2 + 7)x + (3 - 21)y + (1 + 35)z + (20 - 56) = 0$$

$$9x - 18y + 36z - 36 = 0$$

Dividing the entire equation by $9$, we arrive at the simplified equation of the rotated plane:

$$x - 2y + 4z - 4 = 0$$

### Finding the Mirror Image

We seek the reflection $B(a, b, c)$ of point $A(2, -\frac{1}{2}, 2)$ in the plane $x - 2y + 4z - 4 = 0$. The line $AB$ is perpendicular to the plane, and the midpoint of $AB$ lies on the plane.

Using the reflection formula:

$$\frac{a - 2}{1} = \frac{b - (-\frac{1}{2})}{-2} = \frac{c - 2}{4} = \frac{-2(1(2) - 2(-\frac{1}{2}) + 4(2) - 4)}{1^2 + (-2)^2 + 4^2}$$

Evaluating the constant ratio:

$$\frac{-2(2 + 1 + 8 - 4)}{1 + 4 + 16} = \frac{-2(7)}{21} = -\frac{14}{21} = -\frac{2}{3}$$

### Final Calculation

Solving for the coordinates $a, b,$ and $c$:

$$a = 2 - \frac{2}{3} = \frac{4}{3}$$

$$b = -\frac{1}{2} + \frac{4}{3} = \frac{5}{6}$$

$$c = 2 - \frac{8}{3} = -\frac{2}{3}$$

The ratio $a:b:c$ is $\frac{4}{3} : \frac{5}{6} : -\frac{2}{3}$. Multiplying by $6$ to clear the denominators, we get $8 : 5 : -4$.

The final result is confirmed by the relation:

$$\frac{a}{8} = \frac{b}{5} = \frac{c}{-4}$$

Select Answer:

Visualized Solution

Identify the Given Planes

Family of Planes

Group the Variables

The Perpendicularity Condition

Apply Dot Product

Solve for $λ$

Equation of Rotated Plane $P_{3}$

Point A and its Mirror Image

Mirror Image Formula

Substitute Values into Formula

Evaluate the RHS

Solve for $a, b, c$

Determine the Final Ratio

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Perpendicularity Condition

The Rotated Plane

Finding the Mirror Image

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Identify the Given Planes

Family of Planes

Group the Variables

The Perpendicularity Condition

Apply Dot Product

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

Equation of Rotated Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P3​

Point A and its Mirror Image

Mirror Image Formula

Substitute Values into Formula

Evaluate the RHS

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

Determine the Final Ratio

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Perpendicularity Condition

The Rotated Plane

Finding the Mirror Image

Final Calculation

Similar Questions

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A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

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