Last updated at May 29, 2018 by Teachoo

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Misc 18 (Method 1) Find the distance of the point (–1, –5, –10) from the point of intersection of the line 𝑟 = 2 𝑖 – 𝑗 + 2 𝑘 + 𝜆 (3 𝑖 + 4 𝑗 + 2 𝑘) and the plane 𝑟 . ( 𝑖 – 𝑗 + 𝑘) = 5 . Given, the equation of line is 𝑟 = (2 𝑖 − 𝑗 + 2 𝑘) + 𝜆 (3 𝑖 + 4 𝑗 + 2 𝑘) and the equation of the plane is 𝑟.( 𝑖 − 𝑗 + 𝑘) = 5 To find point of intersection of line and plane, putting value of 𝒓 from equation of line into equation of plane. (2 𝑖 − 𝑗 + 2 𝑘) + 𝜆 (3 𝑖 + 4 𝑗 + 2 𝑘) . ( 𝑖 − 𝑗 + 𝑘) = 5 (2 𝑖 − 1 𝑗 + 2 𝑘+3𝜆 𝑖 + 4𝜆 𝑗+2𝜆 𝑘) . (1 𝑖 − 1 𝑗 + 1 𝑘) = 5 (2 + 3𝜆) 𝑖 + (−1 + 4𝜆) 𝑗+(2+2𝜆) 𝑘 . (1 𝑖 − 1 𝑗 + 1 𝑘) = 5 (2 + 3𝜆) × 1 + (−1 + 4𝜆) × (−1) + (2 + 2𝜆) × 1 = 5 2 + 3𝜆 + 1 − 4𝜆 + 2 + 2𝜆 = 5 𝜆 + 5 = 5 𝜆 = 5 − 5 𝜆 = 0 So, the equation of line is 𝑟 = (2 𝑖 − 𝑗 + 2 𝑘) + 𝜆 (3 𝑖 + 4 𝑗 + 2 𝑘) 𝒓 = 2 𝒊 − 𝒋 + 2 𝒌 Let the point of intersection be (x, y, z) So, 𝑟 = x 𝑖 + y 𝑗 + z 𝑘 x 𝑖 + y 𝑗 + z 𝑘 = 2 𝑖 − 𝑗 + 2 𝑘 Hence, x = 2 , y = −1, z = 2 Therefore, the point of intersection is (2, −1, 2) Now, the distance between two points ( 𝑥1, 𝑦1, 𝑧1) and ( 𝑥2, 𝑦2, 𝑧2) is 𝑥2− 𝑥12 + 𝑦2− 𝑦12+ 𝑧2− 𝑧12 Distance between (2, −1, 2) and (−1, −5, −10) = −1−22 + −5+12+ −10−22 = −32 + −42+ −122 = 9+16+144 = 169 = 13. Misc 18 (Method 2) Find the distance of the point (–1, –5, –10) from the point of intersection of the line 𝑟 = 2 𝑖 – 𝑗 + 2 𝑘 + 𝜆 (3 𝑖 + 4 𝑗 + 2 𝑘) and the plane 𝑟 . ( 𝑖 – 𝑗 + 𝑘) = 5 . Given, the equation of line is 𝑟 = (2 𝑖 − 𝑗 + 2 𝑘) + 𝜆 (3 𝑖 + 4 𝑗 + 2 𝑘) Comparing with 𝒓 = 𝒂 + 𝜆 𝒃 , Equation of line in Cartesian form is 𝑥 − 𝑥1𝑎 = 𝑦 − 𝑦1𝑏 = 𝑧 − 𝑧1𝑐 𝑥 − 23 = 𝑦 − (−1)4 = 𝑧 − 22 𝒙 − 𝟐𝟑 = 𝒚 + 𝟏𝟒 = 𝒛 − 𝟐𝟐 = k So, Also, the equation of plane is 𝑟.( 𝑖 − 𝑗 + 𝑘) = 5 Comparing with 𝑟. 𝑛 = d, 𝑛 = 1 𝑖 − 1 𝑗 + 1 𝑘 & d = 5 Comparing 𝑛 with A 𝑖 + B 𝑗 + C 𝑘, A = 1, B = −1, C = 1 Equation of plane in Cartesian form is Ax + By + Cz = d 1x − 1y + 1z = 5 x − y + z = 5 Let the point of intersection of line and plane be (x, y, z) Putting values of x, y, z in equation of plane, (3k + 2) − (4k − 1) + (2k + 2) = 5 3k + 2 − 4k + 1 + 2k + 2 = 5 k + 5 = 5 ∴ k = 0 So, x = 3k + 2 = 3 × 0 + 2 = 2 y = 4k − 1 = 4 × 0 − 1 = −1 z = 2k + 2 = 2 × 0 + 2 = 2 Therefore, the point of intersection is (2, −1, 2). Distance between two points ( 𝑥1, 𝑦1, 𝑧1) & ( 𝑥2, 𝑦2, 𝑧2) = 𝑥2− 𝑥12 𝑦2− 𝑦12+ 𝑧2− 𝑧12 ∴ Distance between (2, −1, 2) and (−1, −5, −10) = 1−22+ −5+12+ −10−22 = −32+ −42+ −122 = 9+16+144 = 169 = 13

Chapter 11 Class 12 Three Dimensional Geometry

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

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 You are here

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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.