Question

The equation of the plane that contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and is perpendicular to the plane $5x+3y-6z+8=0$ is

• A. $14x+7y-7z-4=0$
• B. $33x+45y+50z-41=0$
• C. $-33x+45y-50z+41=0$
• D. $5x+31y+50z-41=0$

Hint:

To avoid tricky fractions, multiply the entire plane equation by the denominator of $\lambda$ (which is 19 here) immediately after substituting it back in.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the equation of a plane that passes through the intersection line of two given planes and is perpendicular to a third given plane.

Step 2: Key Formula or Approach:
The equation of a family of planes passing through the intersection of $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0$. If two planes are perpendicular, the dot product of their normal vectors is zero.

Step 3: Detailed Explanation:
Write the equation of the required plane using the family of planes formula:
$$(x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0$$ Group the $x, y, z$ terms to find the direction ratios of its normal:
$$(1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z + (-4 + 5\lambda) = 0$$ The normal vector to this plane is $\vec{n}_1 = \langle 1 + 2\lambda, 2 + \lambda, 3 - \lambda \rangle$.
This plane is perpendicular to $5x + 3y - 6z + 8 = 0$, whose normal is $\vec{n}_2 = \langle 5, 3, -6 \rangle$.
Apply the perpendicularity condition $\vec{n}_1 \cdot \vec{n}_2 = 0$:
$$5(1 + 2\lambda) + 3(2 + \lambda) - 6(3 - \lambda) = 0$$ $$5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$$ Combine like terms:
$$19\lambda - 7 = 0 \implies \lambda = \frac{7}{19}$$ Substitute $\lambda$ back into the grouped family of planes equation:
$$19(x + 2y + 3z - 4) + 7(2x + y - z + 5) = 0$$ $$19x + 38y + 57z - 76 + 14x + 7y - 7z + 35 = 0$$ $$33x + 45y + 50z - 41 = 0$$

Step 4: Final Answer:
The equation of the plane is $33x + 45y + 50z - 41 = 0$, matching option (B).