Last updated at May 29, 2018 by Teachoo

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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

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Chapter 11 Class 12 Three Dimensional Geometry

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.