Last updated at Dec. 8, 2016 by Teachoo

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Ex 11.3, 6 (Introduction) Find the equations of the planes that passes through three points. (a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3) Vector equation of a plane passing through three points with position vectors 𝑎, 𝑏, 𝑐 is ( r − 𝑎) . ( 𝑏− 𝑎)×( 𝑐− 𝑎) = 0 Ex 11.3, 6 Find the equations of the planes that passes through three points. (a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3) Vector equation of a plane passing through three points with position vectors 𝑎, 𝑏, 𝑐 is ( r − 𝒂) . ( 𝒃− 𝒂)×( 𝒄− 𝒂) = 0 Now, the plane passes through the points ( 𝒃 − 𝒂) = (6 𝑖 + 4 𝑗 – 5 𝑘) − (1 𝑖 + 1 𝑗 − 1 𝑘) = (6 −1) 𝑖 + (4 − 1) 𝑗 + (−5 − (− 1)) 𝑘 = 5 𝒊 + 3 𝒋 − 4 𝒌 ( 𝒄 − 𝒂) = (− 4 𝑖 − 2 𝑗 + 3 𝑘) − (1 𝑖 + 1 𝑗 − 1 𝑘) = (−4 − 1) 𝑖 +(−2 −1) 𝑗 + (3 − (− 1)) 𝑘 = −5 𝒊 − 3 𝒋 + 4 𝒌 ( 𝒃 − 𝒂) × ( 𝒄 − 𝒂) = 𝑖 𝑗 𝑘53 − 4 − 5 − 34 = – 𝑖 𝑗 𝑘53 − 4 5 3− 4 = 𝟎 This implies, the three points are collinear. Ex 11.3, 6 Find the equations of the planes that passes through three points. (b) (1, 1, 0), (1, 2, 1), (– 2, 2, – 1) Vector equation of a plane passing through three points with position vectors 𝑎, 𝑏, 𝑐 is ( r − 𝒂) . ( 𝒃− 𝒂)×( 𝒄 − 𝒂) = 0 Now, the plane passes through the points ( 𝒄 − 𝒂) = (–2 𝑖 + 2 𝑗 + 1 𝑘) − (1 𝑖 + 1 𝑗 +0 𝑘) = (−2 − 1) 𝑖 + (2 − 1) 𝑗 + (−1 − 0) 𝑘 = −3 𝒊 + 1 𝒋 − 1 𝒌 ( 𝒃 − 𝒂) × ( 𝒄 − 𝒂) = 𝑖 𝑗 𝑘011 − 31 − 1 = 𝑖 1 ×−1−(1×1) – 𝑗 0 ×−1−(−3 ×1) + 𝑘 0 ×1−(−3 ×1) = 𝑖(–1 – 1) – 𝑗 (0 + 3) + 𝑘 ( 0 + 3) = –2 𝒊 – 3 𝒋 + 3 𝒌 ∴ Vector equation of plane is 𝑟− 1 𝑖+1 𝑗+0 𝑘 . −2 𝑖−3 𝑗+3 𝑘 = 0 𝒓− 𝒊+ 𝒋 . −𝟐 𝒊−𝟑 𝒋+𝟑 𝒌 = 𝟎 Finding Cartesian equation Put 𝒓 = x 𝒊 + y 𝒋 + z 𝒌 𝑟− 𝑖+ 𝑗 . −2 𝑖−3 𝑗+3 𝑘 = 0 𝑥 𝑖+𝑦 𝑗+𝑧 𝑘−( 𝑖+ 𝑗). −2 𝑖−3 𝑗+3 𝑘 = 0 𝑥−1 𝑖 + 𝑦−1 𝑗+𝑧 𝑘. −2 𝑖−3 𝑗+3 𝑘 = 0 –2(x − 1) + (−3)(y − 1) + 3(z) = 0 –2x + 2 − 3y + 3 + 3z = 0 2x + 3y – 3z = 5 ∴ Equation of plane in Cartesian form is 2x + 3y – 3z = 5

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.