Ex 11.3, 6

Transcript

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﷯﷮ 𝑘﷯﷮5﷮3﷮ − 4﷮ − 5﷮ − 3﷮4﷯﷯ = – 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮5﷮3﷮ − 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﷯﷮ 𝑘﷯﷮0﷮1﷮1﷮ − 3﷮1﷮ − 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