<p>If \(L_1\) is the line of intersection of the planes \(2x - 2y + 3z - 2 = 0\), \(x - y + z + 1 = 0\) and \(L_2\) is the line of intersection of the planes \(x + 2y - z - 3 = 0\), \(3x - y + 2z - 1 = 0\), then the distance of the origin from the plane, containing the lines \(L_1\) and \(L_2\) is</p>
<p>\(\dfrac{1}{3\sqrt{2}}\)</p>
<p>\(\dfrac{1}{2\sqrt{2}}\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
<p>\(\dfrac{1}{4\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: The plane containing two skew/intersecting lines can be found using two direction vectors (from each line) and a point on one line; the normal to this plane is perpendicular to both direction vectors.
Step 1: Find direction vector of L_1 using normal vectors of the two planes. Plane 1: 2x - 2y + 3z - 2 = 0 has normal n_1 = (2, -2, 3) Plane 2: x - y + z + 1 = 0 has normal n_2 = (1, -1, 1) Direction of L_1: d_1 = n_1 × n_2 = (2, -2, 3) × (1, -1, 1) = (1, 1, 0) Step 2: Find direction vector of L_2 using normal vectors of its planes. Plane 3: x + 2y - z - 3 = 0 has normal n_3 = (1, 2, -1) Plane 4: 3x - y + 2z - 1 = 0 has normal n_4 = (3, -1, 2) Direction of L_2: d_2 = n_3 × n_4 = (1, 2, -1) × (3, -1, 2) = (3, -5, -7) Step 3: Find a point on L_1 by setting z = 0 in both planes: 2x - 2y = 2 and x - y = -1 gives point P_1 = (0, 1, 0) Step 4: The normal to the plane containing both lines is: N = d_1 × d_2 = (1, 1, 0) × (3, -5, -7) = (-7, 7, -8) Step 5: Equation of plane: -7x + 7y - 8z = d. Using point (0, 1, 0): d = 7 Plane equation: -7x + 7y - 8z - 7 = 0 or 7x - 7y + 8z + 7 = 0 Step 6: Distance from origin = |7(0) - 7(0) + 8(0) + 7|/√(49 + 49 + 64) = 7/√162 = 7/(9√2) = 7√2/18 ∴ Answer: A
Correct Answer: A