Misc 17

Transcript

Misc 17 Find the equation of the plane which contains the line of intersection of the planes 𝑟﷯ . ( 𝑖﷯ + 2 𝑗﷯ + 3 𝑘﷯) – 4 = 0 , 𝑟﷯ . (2 𝑖﷯ + 𝑗﷯ – 𝑘﷯) + 5 = 0 and which is perpendicular to the plane 𝑟﷯ . (5 𝑖﷯ + 3 𝑗﷯ – 6 𝑘﷯) + 8 = 0 . Equation of a plane passing through the intersection of the places A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is (A1x + B1y + C1z − d1) + 𝜆 (A2x + B2y + C2z – d2) = 0 Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 Equation of plane is (A1x + B1y + C1z − d1) + 𝜆 (A2x + B2y + C2z = d2) = 0 Putting values (1x + 2y + 3z − 4) + 𝜆 ( − 2x − 1y + 1z − 5) = 0 (1 − 2𝜆) x + (2 − 𝜆)y + (3 + 𝜆) z + ( −4 − 5𝜆) = 0 Now, the plane is perpendicular to the plane 𝑟﷯.(5 𝑖﷯ + 3 𝑗﷯ − 6 𝑘﷯) + 8 = 0 So, normal to plane 𝑁﷯ will be perpendicular to normal 𝑛﷯ of 𝑟﷯.(5 𝑖﷯ + 3 𝑗﷯ − 6 𝑘﷯) + 8 = 0 Now, 𝑟﷯.(5 𝑖﷯ + 3 𝑗﷯ − 6 𝑘﷯) + 8 = 0 𝑟﷯ .(5 𝑖﷯ + 3 𝑗﷯ − 6 𝑘﷯) = –8 − 𝑟﷯ .(5 𝑖﷯ + 3 𝑗﷯ − 6 𝑘﷯) = 8 𝑟﷯ .( −5 𝑖﷯ − 3 𝑗﷯ + 6 𝑘﷯) = 8 Finding direction cosines of 𝑁﷯ & 𝑛﷯ Since, 𝑁﷯ is perpendicular to 𝑛﷯ 𝑎1 𝑎2 + b1 b2 + c1 c2 = 0 (1 − 2𝜆) × −5 + (2 − 𝜆) × −3 + (3 + 𝜆) × 6 = 0 − 5 + 10𝜆 − 6 + 3𝜆 + 18 + 6𝜆 = 0 19𝜆 + 7 = 0 ∴ 𝜆 = −𝟕﷮𝟏𝟗﷯ Putting value of 𝜆 in (1), (1 − 2𝜆) x + (2 − 𝜆)y + (3 + 𝜆) z + ( −4 − 5𝜆) = 0 1−2 × −7﷮19﷯﷯ x + 2− −7﷮19﷯﷯﷯y + 3+ − 7﷮19﷯﷯﷯z + −4−5× −7﷮19﷯﷯ = 0 1 + 14﷮19﷯﷯ x + 2 + 7﷮19﷯﷯y + 3 − 7﷮19﷯﷯z + − 4 + 35﷮19﷯﷯ = 0 33﷮19﷯ x + 45﷮19﷯ y + 50﷮19﷯ z − 41﷮19﷯ = 0 1﷮19﷯ (33x + 45y + 50z − 41) = 0 33x + 45y + 50z − 41 = 0 Therefore, the equation of the plane is 33x + 45y + 50z = 41.