Determine the vector equation of the plane passing through the intersection of the planes r × (i + j + k) = 6 and r × (2i + 3j + 4k) = -5 , and throug - Mathematics MCQ Mathematics Mock Test - 8 2026 | ExamTest.live

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Determine the vector equation of the plane passing through the intersection of the planes r \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 and r \cdot (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = - 5, and through the point (1, 1, 1) ?

Solution & Step-by-step Explanation

The equation of a plane passing through the intersection of two planes P_{1} and P_{2} is P_{1} + λ P_{2} = 0 .In Cartesian form: P_{1} : x + y + z - 6 = 0 P_{2} : 2 x + 3 y + 4 z + 5 = 0 The family of planes is: (x + y + z - 6) + λ (2 x + 3 y + 4 z + 5) = 0 Since it passes through (1, 1, 1) : (1 + 1 + 1 - 6) + λ (2 (1) + 3 (1) + 4 (1) + 5) = 0 - 3 + λ (14) = 0 ⟹ λ = \frac{3}{14} Substitute λ back: (x + y + z - 6) + \frac{3}{14} (2 x + 3 y + 4 z + 5) = 0 Multiply by 14: 14 x + 14 y + 14 z - 84 + 6 x + 9 y + 12 z + 15 = 0 20 x + 23 y + 26 z - 69 = 0 20 x + 23 y + 26 z = 69 In vector form: r \cdot (20 \hat{i} + 23 \hat{j} + 26 \hat{k}) = 69

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Determine the vector equation of the plane passing through the intersection of the planes r \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 and r \cdot (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = - 5, and through the point (1, 1, 1) ?

r \cdot (10 \hat{i} + 13 \hat{j} + 23 \hat{k}) = 69

r \cdot (20 \hat{i} + 23 \hat{j} + 26 \hat{k}) = 69

r \cdot (30 \hat{i} + 23 \hat{j} + 13 \hat{k}) = 69

r \cdot (10 \hat{i} + 23 \hat{j} + 13 \hat{k}) = 69

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