Last updated at May 29, 2018 by Teachoo

Transcript

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

Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important You are here

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23 Important

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

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.