<p><strong>Paragraph for Question nos. 640 and 641</strong><br>Consider two lines \(L_1: 2\hat{i}+\hat{j}-\hat{k}+\lambda(\hat{i}+2\hat{k})\) and \(L_2: 3\hat{i}+\hat{j}-\hat{k}+\mu(\hat{i}+\hat{j}-\hat{k})\). Let \(\Pi\) be the plane which contains the line \(L_1\) and parallel to \(L_2\) and intersecting coordinate axes at \(A\), \(B\) and \(C\) respectively.</p><p>Volume of tetrahedron \(OABC\), (where \(O\) is origin) is:</p>
<p>(a) \(\dfrac{2}{3}\)</p>
<p>(b) \(\dfrac{4}{9}\)</p>
<p>(c) \(\dfrac{2}{9}\)</p>
<p>(d) \(\dfrac{4}{3}\)</p>
Step-by-Step Solution
Key Concept: A plane containing line L₁ and parallel to L₂ must have a normal vector perpendicular to both direction vectors. Use the cross product of L₁'s direction and L₂'s direction to find the plane equation, then find intercepts to calculate the tetrahedron volume.
Step 1: Find the normal vector to plane Π L_1 has direction vector d_1 = (1, 0, 2) L_2 has direction vector d_2 = (1, 1, -1) Normal to Π: n = d_1 × d_2 = | i j k | = (1, 0, 2) × (1, 1, -1) = (-2, 3, 1) Step 2: Find plane equation using point on L_1 When λ = 0, point on L_1 is P = (2, 1, -1) Plane equation: -2(x-2) + 3(y-1) + 1(z+1) = 0 -2x + 4 + 3y - 3 + z + 1 = 0 -2x + 3y + z + 2 = 0 or 2x - 3y - z = 2 Step 3: Find intercepts on coordinate axes A (x-intercept): Set y=0, z=0 → 2x = 2 → x = 1, so A = (1, 0, 0) B (y-intercept): Set x=0, z=0 → -3y = 2 → y = -2/3, so B = (0, -2/3, 0) C (z-intercept): Set x=0, y=0 → -z = 2 → z = -2, so C = (0, 0, -2) Step 4: Calculate volume of tetrahedron OABC V = (1/6)|OA · (OB × OC)| = (1/6)|1 × (-2/3) × (-2)| V = (1/6)|4/3| = 4/18 = 2/9 ∴ Answer: D
Correct Answer: D