# Example 27

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 27 (Method 1) Find the equation of the plane that contains the point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. 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, −1, 2) So, equation of plane is A(x −1) + B (y + 1) + 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 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Required normal = 𝑖 𝑗 𝑘23−212−3 = 𝑖 (3(–3) – 2(–2)) – 𝑗 (2(–3) – 1(–2)) + 𝑘(2(2) – 1(3)) = 𝑖 (–9 + 4) – 𝑗 (–6 + 2) + 𝑘(4 – 3) = –5 𝑖 + 4 𝑗 + 𝑘 Hence, direction ratios = –5, 4, 1 ∴ A = –5, B = 4, C = 1 Putting above values in (1), A(x −1) + B (y + 1) + C(z − 2) = 0 −5(x − 1) + 4 (y + 1) + 1 (z − 2) = 0 −5x + 5 + 4y + 4 + z − 2 = 0 −5x + 4y + z + 7 = 0 −5x + 4y + z = −7 −(5x −4y − z) = −7 5x − 4y − z = 7 Therefore, the equation of the required plane is 5x − 4y − z = 7. Example 27 (Method 2) Find the equation of the plane that contains the point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. 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, −1, 2) So, equation of plane is A(x −1) + B (y + 1) + 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 + 1) + C(z − 2) = 0 is perpendicular to plane 2x + 3y – 2z = 5 Hence, A × 2 + B × 3 + C × (–2) = 0 2A + 3B − 2C = 0 Similarly, Given that plane A(x −1) + B (y + 1) + C(z − 2) = 0 is perpendicular to plane x + 2y – 3z = 8 Hence, A × 1 + B × 2 + C × (–3) = 0 A + 2B − 3C = 0 So, our equations are 2A + 3B −2C = 0 …(2) A + 2B − 3C = 0 …(3) Solving 𝐴−9 − (−4) = 𝐵−2 − (−6) = 𝐶4 − 3 𝐴−9 + 4 = 𝐵−2 + 6 = 𝐶1 𝐴−5 = 𝐵4 = 𝐶1 = k So, A = –5k , B = 4k , C = k Putting above values in (1), A(x −1) + B (y + 1) + C(z − 2) = 0 −5k(x − 1) + 4k (y + 1) + k (z − 2) = 0 k[−5(x − 1) + 4(y + 1) + (z − 2)] = 0 −5x + 5 + 4y + 4 + z − 2 = 0 −5x + 4y + z + 7 = 0 −5x + 4y + z = −7 −(5x −4y − z) = −7 5x − 4y − z = 7 Therefore, the equation of the required plane is 5x − 4y − z = 7.

Example, 3
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Example 27 Important You are here

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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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 .