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

Serial order wise

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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.