The plane $3x + 4y + 12z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $5x + 3y + 10z = 25$. If the equation of the plane in new position is $x - 4y + 6x = K$, then the value of $K$ is
Step-by-Step Solution
Key Concept: For two planes to be parallel, their normal vectors must be proportional; use this condition to find the parameter.
The plane through the line of intersection is $(4x + 7y + 4z + 81) + \lambda(5x + 3y + 10z - 25) = 0$. Expanding: $(4 + 5\lambda)x + (7 + 3\lambda)y + (4 + 10\lambda)z + (81 - 25\lambda) = 0$. For this plane to be parallel to $x - 4y + 6z = K$, the normal vectors must be proportional: $\frac{4 + 5\lambda}{1} = \frac{7 + 3\lambda}{-4} = \frac{4 + 10\lambda}{6}$. From the first two ratios: $-4(4 + 5\lambda) = 7 + 3\lambda$, giving $-16 - 20\lambda = 7 + 3\lambda$, so $\lambda = -1$. The normal becomes $(−1, 4, −6)$ or equivalently $(1, −4, 6)$. Using the third part to find $K$: $K = 106$.
Correct Answer: 106