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Misc 13 - Equation of plane passing (-1, 3, 2) perpendicular - Angle between two planes

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Misc 13 (Method 1) Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. The equation of a plane passing through ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) is given by A(x − 𝒙﷮𝟏﷯) + B (y − 𝒚﷮𝟏﷯) + C(z – 𝒛﷮𝟏﷯) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (–1, 3, 2) So, equation of plane is A(x + 1) + B (y – 3) + C(z − 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C Also, the plane is perpendicular to the given two planes. So, their normal to plane would be perpendicular to normals of both planes. We know that 𝑎﷯ × 𝑏﷯ is perpendicular to both 𝑎﷯ & 𝑏﷯ So, required is normal is cross product of normals of planes x + 2y + 3z = 5 and 3x + 3y + z = 0 Required normal = 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮1﷮2﷮3﷮3﷮3﷮1﷯﷯ = 𝑖﷯ (2(1) – 3(3)) – 𝑗﷯ (1(1) – 3(3)) + 𝑘﷯(1(3) – 3(2)) = 𝑖﷯ (2 – 9) – 𝑗﷯ (1 – 9) + 𝑘﷯(3 – 6) = –7 𝑖﷯ + 8 𝑗﷯ – 3 𝑘﷯ Hence, direction ratios = –7, 8, –3 ∴ A = –7, B = 8, C = –3 Putting above values in (1) A(x + 1) + B (y – 3) + C(z − 2) = 0 −7k (x + 1) + 8k (y − 3) − 3k (z −2) = 0 k −7 𝑥+1﷯+8 𝑦−3﷯−3(𝑧−2)﷯ = 0 −7(x + 1) + 8(y − 3) − 3(z − 2) = 0 −7x − 7 + 8y − 24 − 3z + 6 = 0 −7x + 8y −3z − 25 = 0 0 = 7x – 8y + 3z + 25 7x – 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x – 8y + 3z + 25 = 0 Misc 13 (Method 2) Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. The equation of a plane passing through ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) is given by A(x − 𝒙﷮𝟏﷯) + B (y − 𝒚﷮𝟏﷯) + C(z – 𝒛﷮𝟏﷯) = 0 where, A, B, C are the direction ratios of normal to the plane. Now the plane passes through (–1, 3, 2) So, equation of plane is A(x + 1) + B (y – 3) + C(z − 2) = 0 We find the direction ratios of normal to plane i.e. A, B, C Also, the plane is perpendicular to the given two planes. Now, it is given that plane A(x + 1) + B (y – 3) + C(z − 2) = 0 is perpendicular to plane x + 2y + 3z = 5 Hence, A × 1 + B × 2 + C × 3 = 0 A + 2B + 3C = 0 Similarly, Given that plane A(x + 1) + B (y – 3) + C(z − 2) = 0 is perpendicular to plane 3x + 3y + z = 0. Hence, A × 3 + B × 3 + C × 1 = 0 3A + 3B + C = 0 So, our equations are A + 2B + 3C = 0 3A + 3B + C = 0 Solving 𝐴﷮2 − 9﷯ = 𝐵﷮9 − 1﷯ = 𝐶﷮3 − 6﷯ 𝑨﷮−𝟕﷯ = 𝑩﷮𝟖﷯ = 𝑪﷮−𝟑﷯ = k So, A = −7k, B = 8k, C = −3k Putting above values in (1) A(x + 1) + B (y – 3) + C(z − 2) = 0 −7k (x + 1) + 8k (y − 3) − 3k (z −2) = 0 k −7 𝑥+1﷯+8 𝑦−3﷯−3(𝑧−2)﷯ = 0 −7(x + 1) + 8(y − 3) − 3(z − 2) = 0 −7x − 7 + 8y − 24 − 3z + 6 = 0 −7x + 8y −3z − 25 = 0 0 = 7x – 8y + 3z + 25 7x – 8y + 3z + 25 = 0 Therefore equation of the required plane is 7x – 8y + 3z + 25 = 0

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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.
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