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Last updated at Feb. 4, 2020 by Teachoo
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Misc 18 (Method 1) Find the distance of the point (โ1, โ5, โ10) from the point of intersection of the line ๐ โ = 2๐ ฬ โ ๐ ฬ + 2๐ ฬ + ๐ (3๐ ฬ + 4๐ ฬ + 2๐ ฬ) and the plane ๐ โ . (๐ ฬ โ ๐ ฬ + ๐ ฬ) = 5 .Given, the equation of line is ๐ โ = (2๐ ฬ โ ๐ ฬ + 2๐ ฬ) + ๐ (3๐ ฬ + 4๐ ฬ + 2๐ ฬ) and the equation of the plane is ๐ โ.(๐ ฬ โ ๐ ฬ + ๐ ฬ) = 5 To find point of intersection of line and plane, Putting value of ๐ โ from equation of line into equation of plane. ["(2" ๐ ฬ" โ " ๐ ฬ" + 2" ๐ ฬ") + ๐ (3" ๐ ฬ" + 4" ๐ ฬ" + 2" ๐ ฬ")" ] . (๐ ฬ โ ๐ ฬ + ๐ ฬ) = 5 ["(2" ๐ ฬ" โ 1" ๐ ฬ" + 2" ๐ ฬ+3"๐" ๐ ฬ" + 4๐" ๐ ฬ+2"๐" ๐ ฬ")" ] . (1๐ ฬ โ 1๐ ฬ + 1๐ ฬ) = 5 ["(2 + 3๐) " ๐ ฬ" + (" โ1" + 4๐) " ๐ ฬ+(2+2"๐" )๐ ฬ ] . (1๐ ฬ โ 1๐ ฬ + 1๐ ฬ) = 5 (2 + 3๐) ร 1 + (โ1 + 4๐) ร (โ1) + (2 + 2๐) ร 1 = 5 2 + 3๐ + 1 โ 4๐ + 2 + 2๐ = 5 ๐ + 5 = 5 ๐ = 5 โ 5 ๐ = 0 So, the equation of line is ๐ โ = (2๐ ฬ โ ๐ ฬ + 2๐ ฬ) + ๐ (3๐ ฬ + 4๐ ฬ + 2๐ ฬ) ๐ โ = 2๐ ฬ โ ๐ ฬ + 2๐ ฬ Let the point of intersection be (x, y, z) So, ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ x๐ ฬ + y๐ ฬ + z๐ ฬ = 2๐ ฬ โ ๐ ฬ + 2๐ ฬ Hence, x = 2 , y = โ1, z = 2 Therefore, the point of intersection is (2, โ1, 2) Now, the distance between two points (๐ฅ_1, ๐ฆ_1, ๐ง_1) and (๐ฅ_2, ๐ฆ_2, ๐ง_2) is โ((๐ฅ_2โ๐ฅ_1 )^2 ใ+ (๐ฆ_2โ๐ฆ_1 )ใ^2+ (๐ง_2โ๐ง_1 )^2 ) Distance between (2, โ1, 2) and (โ1, โ5, โ10) = โ((โ1โ2)^2 ใ+ (โ5+1)ใ^2+ (โ10โ2)^2 ) = โ((โ3)^2 ใ+ (โ4)ใ^2+ (โ12)^2 ) = โ(9+16+144) = โ169 = 13. Misc 18 (Method 2) Find the distance of the point (โ1, โ5, โ10) from the point of intersection of the line ๐ โ = 2๐ ฬ โ ๐ ฬ + 2๐ ฬ + ๐ (3๐ ฬ + 4๐ ฬ + 2๐ ฬ) and the plane ๐ โ . (๐ ฬ โ ๐ ฬ + ๐ ฬ) = 5 .Given, the equation of line is ๐ โ = (2๐ ฬ โ ๐ ฬ + 2๐ ฬ) + ๐ (3๐ ฬ + 4๐ ฬ + 2๐ ฬ) Comparing with ๐ โ = ๐ โ + ๐๐ โ , ๐ โ = 2๐ ฬ โ ๐ ฬ + 2๐ ฬ Comparing with ๐ โ = ๐ฅ_1 ๐ ฬ + ๐ฆ_1 ๐ ฬ + ๐ง_1 ๐ ฬ, โด ๐ฅ_1= 2, ๐ฆ_1= โ1, ๐ง_1= 2, ๐ โ = 3๐ ฬ + 4๐ ฬ + 2๐ ฬ Comparing with ๐ โ = ๐๐ ฬ + ๐๐ ฬ + ๐๐ ฬ, โด ๐ = 3, ๐ = 4, ๐ = 2, Equation of line in Cartesian form is (๐ฅ โ ๐ฅ_1)/๐ = (๐ฆ โ ๐ฆ_1)/๐ = (๐ง โ ๐ง_1)/๐ (๐ฅ โ 2)/3 = (๐ฆ โ (โ1))/4 = (๐ง โ 2)/2 (๐ โ ๐)/๐ = (๐ + ๐)/๐ = (๐ โ ๐)/๐ = k So, Also, the equation of plane is ๐ โ.(๐ ฬ โ ๐ ฬ + ๐ ฬ) = 5 Comparing with ๐ โ.๐ โ = d, ๐ โ = 1๐ ฬ โ 1๐ ฬ + 1๐ ฬ & d = 5 Comparing ๐ โ with A๐ ฬ + B๐ ฬ + C๐ ฬ, A = 1, B = โ1, C = 1 Equation of plane in Cartesian form is Ax + By + Cz = d 1x โ 1y + 1z = 5 x โ 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) โ (4k โ 1) + (2k + 2) = 5 3k + 2 โ 4k + 1 + 2k + 2 = 5 k + 5 = 5 โด k = 0 So, x = 3k + 2 = 3 ร 0 + 2 = 2 y = 4k โ 1 = 4 ร 0 โ 1 = โ1 z = 2k + 2 = 2 ร 0 + 2 = 2 Therefore, the point of intersection is (2, โ1, 2). Distance between two points (๐ฅ_1, ๐ฆ_1, ๐ง_1) & (๐ฅ_2, ๐ฆ_2, ๐ง_2) = โ((๐ฅ_2โ๐ฅ_1 )^2 (๐ฆ_2โ๐ฆ_1 )^2+(๐ง_2โ๐ง_1 )^2 ) โด Distance between (2, โ1, 2) and (โ1, โ5, โ10) = โ((โ1โ2)^2+(โ5+1)^2+(โ10โ2)^2 ) = โ((โ3)^2+(โ4)^2+(โ12)^2 ) = โ(9+16+144) = โ169 = 13
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