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Find the equation of the plane through the inter section of planes 3x – y + 2z – 4 = 0 and x + y - z - 2 = 0 and the point (2, 2, 1). |
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Answer» Equation of any plane through the intersection of the given planes, is of the form 3x - y + 2z - 4 + λ (x + y - z - 2) = 0 (3 + λ)x + (λ - 1)y + (2 - λ) z - (4 + 2λ) = 0 By Data it is passes through (2, 2, 1) ⇒ (3 + λ)2 + (λ -1)2 + (2 – λ)1 – (4 + 2λ) = 0 ⇒ 6 + 2λ + 2λ - 2 + 2 - λ - 4 - 2λ = 0 ⇒ 2 + λ = 0 λ = - 2 Thus the required equation is (3 - 2)x + (-2 -1 )y + (2 - 2)1 - (4 + 2(-2)) = 0 x - 3y + (0)(1) - (4 - 4) = 0 x - 3y = 0 |
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