Last updated at Aug. 25, 2017 by Teachoo

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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 ( 𝑥1, 𝑦1, 𝑧1) is given by A(x − 𝒙𝟏) + B (y − 𝒚𝟏) + C(z – 𝒛𝟏) = 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 𝑘123331 = 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 ( 𝑥1, 𝑦1, 𝑧1) is given by A(x − 𝒙𝟏) + B (y − 𝒚𝟏) + C(z – 𝒛𝟏) = 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 𝐴2 − 9 = 𝐵9 − 1 = 𝐶3 − 6 𝑨−𝟕 = 𝑩𝟖 = 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

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .