3D Geometry
Planes and intersection
Grade 12
Question:
<p>The equation of the plane through the intersection of the planes \(x + y + z = 1\) and \(2x + 3y - z + 4 = 0\) and parallel to X-axis, is</p>
<p>(a) \(y - 3z + 6 = 0\)</p>
<p>(b) \(3y - z + 6 = 0\)</p>
<p>(c) \(y + 3z + 6 = 0\)</p>
<p>(d) \(3y - 2z + 6 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the family of planes through the intersection of two planes, then apply the condition that the plane is parallel to the X-axis by making the coefficient of x zero.
Solution: The equation of the plane through the intersection of the planes \(x + y + z = 1\) and \(2x + 3y - z + 4 = 0\) is: \((x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0\) or, \((2\lambda + 1)x + (3\lambda + 1)y + (1 - \lambda)z + 4\lambda - 1 = 0\) ... (i) It is parallel to X-axis, i.e. direction ratios are \(1, 0, 0\) Therefore, \(1(2\lambda + 1) + 0(3\lambda + 1) + 0(1 - \lambda) = 0\) \(\Rightarrow 2\lambda + 1 = 0\) \(\Rightarrow \lambda = -\frac{1}{2}\) Substituting \(\lambda = -\frac{1}{2}\) in Eq. (i), we get: \(y - 3z + 6 = 0\) ∴ Answer is (a).
Correct Answer: a