3D Geometry
Plane Equations
Grade 12
Question:
<p>Equation of a plane passing through the lines \(2x - y + z - 3 = 0\), \(3x + y + z - 5 = 0\) and which is at a distance of \(\frac{1}{6}\) from the point \((2, 1, -1)\) is</p>
<p>(a) \(2x - y + z - 3 = 0\)</p>
<p>(b) \(3x + y + z - 5 = 0\)</p>
<p>(c) \(62x + 29y + 19z - 105 = 0\)</p>
<p>(d) \(x + 2y - 2 = 0\)</p>
Step-by-Step Solution
Key Concept: Plane through the intersection of two lines can be expressed as a family of planes. Apply the distance condition to find the specific plane(s).
Step 1: Equation of a plane through the given line is \(2x - y + z - 3 + \lambda(3x + y + z - 5) = 0\) Step 2: Simplify: \((2 + 3\lambda)x + (\lambda - 1)y + (\lambda + 1)z - (3 + 5\lambda) = 0\) Step 3: Use distance formula: \(\frac{|2(2 + 3\lambda) + (\lambda - 1) - (\lambda + 1) - 3 - 5\lambda|}{6\sqrt{(2 + 3\lambda)^2 + (\lambda - 1)^2 + (\lambda + 1)^2}} = \frac{1}{6}\) Step 4: This gives \(11\lambda^2 + 12\lambda + 6 = 6(\lambda - 2\lambda + 1)\) Step 5: Simplifying: \(5\lambda^2 + 24\lambda = 0\) Step 6: Therefore \(\lambda = 0\) or \(\lambda = -\frac{24}{5}\) Step 7: The required planes are \(2x - y + z - 3 = 0\) or \(62x + 29y + 19z - 105 = 0\) ∴ Answer is (a) and (c).
Correct Answer: a,c