Question

The equation of the plane passing through the intersection of planes \(2x-y+z = 3\), \(4x-3y+5z = -9\) and parallel to the line \(\frac{x+1}{2} = \frac{y+3}{4} = \frac{z-3}{5}\) is...

• A. \(11x-3y-2z-54 = 0\)
• B. \(11x+3y-2z-54 = 0\)
• C. \(11x-3y+2z-54 = 0\)
• D. \(11x-3y-2z+54 = 0\)

Hint:

Use the family \(P_1+\lambda P_2=0\) and make its normal perpendicular to the line direction.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The required plane belongs to the family \((2x-y+z-3) + \lambda(4x-3y+5z+9) = 0\).

Step 2: Expand:
\[ (2+4\lambda)x + (-1-3\lambda)y + (1+5\lambda)z + (-3+9\lambda) = 0 \]

Step 3: Use the parallel line:
The line has direction \((2,4,5)\). For a plane to be parallel to it, the normal must be perpendicular to \((2,4,5)\).
\(2(2+4\lambda) + 4(-1-3\lambda) + 5(1+5\lambda) = 0\), i.e. \(4+8\lambda-4-12\lambda+5+25\lambda = 0\), so \(5 + 21\lambda = 0\) and \(\lambda = -\frac{5}{21}\).

Step 4: Substitute:
Coefficient of \(x\): \(2 - \frac{20}{21} = \frac{22}{21}\). Of \(y\): \(-1 + \frac{15}{21} = -\frac{6}{21}\). Of \(z\): \(1 - \frac{25}{21} = -\frac{4}{21}\). Constant: \(-3 - \frac{45}{21} = -\frac{108}{21}\).
Multiply by \(21\): \(22x - 6y - 4z - 108 = 0\), i.e. \(11x - 3y - 2z - 54 = 0\).

Final Answer:
The plane is \(11x-3y-2z-54=0\), option (A). \[ \boxed{11x-3y-2z-54=0} \]