# Example 20

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 20 Find the vector equation of the plane passing through the intersection of the planes 𝑟 . ( 𝑖 + 𝑗 + 𝑘) = 6 and 𝑟 . (2 𝑖 + 3 𝑗 + 4 𝑘)= − 5, and the point (1, 1, 1). The vector equation of a plane passing through the intersection of planes 𝑟. 𝑛1 = d1 and 𝑟. 𝑛2 = d2 and also through the point (x1, y1, z1) is 𝒓.( 𝒏𝟏 + 𝜆 𝒏𝟐) = d1 + 𝜆d2 Given, the plane passes through Equation of plane is 𝑟. 𝑖+ 𝑗+ 𝑘+𝜆(−2 𝑖−3 𝑗−4 𝑘) = 6 + 𝜆5 𝒓. 𝒊 + 𝒋 + 𝒌−𝜆(𝟐 𝒊+𝟑 𝒋+𝟒 𝒌) = 6 + 5𝜆 Now to find 𝜆 , put 𝒓 = x 𝒊 + y 𝒋 + z 𝒌 (x 𝑖 + y 𝑗 + z 𝑘). 𝑖+ 𝑗+ 𝑘−𝜆(2 𝑖+3 𝑗+4 𝑘) = 5𝜆 + 6 (x 𝑖 + y 𝑗 + z 𝑘). 𝑖+ 𝑗+ 𝑘 − 𝜆 (x 𝑖 + y 𝑗 + z 𝑘).(2 𝑖+3 𝑗+4 𝑘) = 5𝜆 + 6 (x × 1) + (y × 1) + (z × 1) − 𝜆 𝑥×2+ 𝑦×3+(𝑧×4) = 5𝜆 + 6 x + y + z − 𝜆 2𝑥+3𝑦+4𝑧 = 5𝜆 + 6 x + y + z − 2𝜆𝑥 − 3𝜆y − 4𝜆z = 5𝜆 + 6 (1 − 2𝜆)x + (1 − 3𝜆)y + (1 − 4𝜆) z = 5𝜆 + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 − 2𝜆)x + (1 − 3𝜆)y + (1 − 4𝜆) z = 5𝜆 + 6 (1 −2𝜆) × 1 + (1 − 3𝜆) × 1 + (1 − 4𝜆) × 1 = 5𝜆 + 6 1 −2𝜆 + 1 − 3𝜆 + 1 − 4𝜆= 5𝜆 + 6 3 − 9𝜆 = 5𝜆 + 6 −14𝜆 = 3 ∴ 𝜆 = − 𝟑𝟏𝟒 Putting value of 𝜆 in (1), 𝑟. 𝑖 + 𝑗 + 𝑘− − 314(2 𝑖+3 𝑗+4 𝑘)= 6 + 5 × − 314 𝑟. 𝑖+ 𝑗+ 𝑘+ 314(2 𝑖+3 𝑗+4 𝑘)= 6 − 1514 𝑟. 𝑖+ 𝑗 + 𝑘+ 614 𝑖+ 914 𝑗+ 1214 𝑘= 6914 𝑟. 1+ 614 𝑖 + 1+ 914 𝑗+ 1+ 1214 𝑘= 6914 𝑟. 2014 𝑖 + 2314 𝑗 + 2614 𝑘= 6914 𝑟. 114(20 𝑖+23 𝑗+26 𝑘)= 6914 114 𝑟. (20 𝑖 + 23 𝑗 + 26 𝑘) = 6914 𝑟. (20 𝑖 + 23 𝑗 + 26 𝑘) = 69 Therefore, the vector equation of the required plane is 𝒓.(𝟐𝟎 𝒊 + 𝟐𝟑 𝒋 + 𝟐𝟔 𝒌) = 𝟔𝟗

Example, 3
Important

Ex 11.1, 2 Important

Example, 6 Important

Example, 9 Important

Example 12 Important

Ex 11.2, 5 Important

Ex 11.2, 11 Important

Ex 11.2, 12 Important

Ex 11.2, 14 Important

Ex 11.2, 15 Important

Ex 11.2, 17 Important

Example 20 Important You are here

Example 21 Important

Example 23 Important

Example 24 Important

Example, 25 Important

Ex 11.3, 4 Important

Ex 11.3, 11 Important

Ex 11.3, 12 Important

Ex 11.3, 14 Important

Example 27 Important

Example 29 Important

Example 30 Important

Misc 6 Important

Misc 9 Important

Misc 14 Important

Misc 18 Important

Misc 20 Important

Misc 21 Important

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

About the Author

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 .