<p>The value(s) of \(m\) for which the straight lines \(\vec{r} = 3\vec{i} - 2\vec{j} - 4\vec{k} + \lambda_1(\vec{i} - \vec{j} + m\vec{k})\) and \(\vec{r} = 5\vec{i} - 2\vec{j} + \vec{k} + \lambda_2(\vec{i} + m\vec{j} + 2\vec{k})\) are coplanar is/are:</p>
<p>(a) \(\frac{5 + \sqrt{33}}{4}\)</p>
<p>(b) \(\frac{-5 + \sqrt{33}}{4}\)</p>
<p>(c) \(\frac{5 - \sqrt{33}}{4}\)</p>
<p>(d) \(\frac{-5 - \sqrt{33}}{4}\)</p>
Step-by-Step Solution
Key Concept: Two lines in 3D are coplanar if and only if the scalar triple product of their direction vectors and the vector connecting any point on one line to any point on the other equals zero. This condition is expressed as: (P₂ - P₁) · (d₁ × d₂) = 0.
Step 1: Identify the given lines. Line 1: Point P_1 = (3, -2, -4), Direction d_1 = (1, -1, m) Line 2: Point P_2 = (5, -2, 1), Direction d_2 = (1, m, 2) Step 2: Find the connecting vector. P_2 - P_1 = (5-3, -2-(-2), 1-(-4)) = (2, 0, 5) Step 3: Apply the coplanarity condition. For coplanarity: (P_2 - P_1) · (d_1 × d_2) = 0 Step 4: Calculate d_1 × d_2. d_1 × d_2 = |i j k| |1 -1 m| |1 m 2| = i[(-1)(2) - m·m] - j[(1)(2) - m·1] + k[(1)(m) - (-1)(1)] = i[-2 - m^2] - j[2 - m] + k[m + 1] = (-2 - m^2, -2 + m, m + 1) Step 5: Compute the scalar triple product. (2, 0, 5) · (-2 - m^2, -2 + m, m + 1) = 0 2(-2 - m^2) + 0(-2 + m) + 5(m + 1) = 0 -4 - 2m^2 + 5m + 5 = 0 -2m^2 + 5m + 1 = 0 2m^2 - 5m - 1 = 0 Step 6: Solve the quadratic equation. Using the quadratic formula: m = [5 ± √(25 + 8)]/4 = [5 ± √33]/4 m_1 = (5 + √33)/4 m_2 = (5 - √33)/4 Step 7: Match with options. m_1 = (5 + √33)/4 matches option (a) m_2 = (5 - √33)/4 matches option (c) The quadratic gives two valid solutions. ∴ Answer: b,c
Correct Answer: b,c