Question

Equation of the plane passing through $(1,-1,2)$ and perpendicular to the planes $x+2y-2z=4$ and $3x+2y+z=6$ is

• A. $6x-7y-4z-5=0$
• B. $6x+7y-4z+5=0$
• C. $6x-7y+4z+5=0$
• D. $6x+7y+4z-5=0$

Hint:

Geometry Tip: Instead of solving the linear system, you can directly find the normal vector of the required plane by taking the cross product of the normal vectors of the two given perpendicular planes: $\vec{n} = \vec{n}_1 \times \vec{n}_2 = \langle1, 2, -2\rangle \times \langle3, 2, 1\rangle = \langle6, -7, -4\rangle$.

Answer 1

The Correct Option is A

Concept: 3D Geometry - Equation of a Plane and Cross Product of Normal Vectors.

Step 1: Set up the general equation for the required plane. The equation of any plane passing through a given point $(x_1, y_1, z_1)$ can be written as $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$, where $a, b, c$ are the direction ratios of the plane's normal. For the point $(1, -1, 2)$, the equation is $a(x - 1) + b(y + 1) + c(z - 2) = 0$.

Step 2: Establish the first perpendicularity condition. If two planes are perpendicular, their normal vectors are orthogonal, meaning their dot product is zero ($a_1a_2 + b_1b_2 + c_1c_2 = 0$). Our required plane is perpendicular to $x + 2y - 2z = 4$. The normal vector of this plane is $(1, 2, -2)$. This gives our first condition: $1a + 2b - 2c = 0$.

Step 3: Establish the second perpendicularity condition. Similarly, our required plane is perpendicular to the second plane $3x + 2y + z = 6$, whose normal vector is $(3, 2, 1)$. This gives our second condition: $3a + 2b + 1c = 0$.

Step 4: Solve for the direction ratios using cross-multiplication. We have a system of two homogeneous linear equations: 1) $a + 2b - 2c = 0$ 2) $3a + 2b + c = 0$ We solve for proportional values of $a, b, c$ using cross-multiplication: $\frac{a}{(2)(1) - (-2)(2)} = \frac{b}{(-2)(3) - (1)(1)} = \frac{c}{(1)(2) - (2)(3)}$ $\frac{a}{2 + 4} = \frac{b}{-6 - 1} = \frac{c}{2 - 6}$ $\frac{a}{6} = \frac{b}{-7} = \frac{c}{-4}$. We can assign these proportional values directly to the direction ratios: $a = 6$, $b = -7$, $c = -4$.

Step 5: Substitute the direction ratios back to form the final equation. Substitute $a = 6$, $b = -7$, and $c = -4$ back into the general equation from
Step 1: $6(x - 1) - 7(y + 1) - 4(z - 2) = 0$. Expand the terms: $6x - 6 - 7y - 7 - 4z + 8 = 0$. Combine the constant terms: $6x - 7y - 4z - 5 = 0$. This matches option A. $$ \therefore \text{The equation of the required plane is } 6x-7y-4z-5=0. $$