# Misc 18

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 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

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Example 27 Important

Example 29 Important

Example 30 Important

Misc 6 Important

Misc 9 Important

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Misc 18 Important You are here

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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 .