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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Feb. 1, 2020 by Teachoo

Example 29 Show that the lines (π₯ β π + π)/(πΌ β πΏ) = (π¦ β π)/πΌ = (π§ β π β π)/(πΌ + πΏ) nd (π₯ β π + π)/(π½ β πΎ) = (π¦ β π)/π½ = (π§ β π β π)/(π½ + πΎ) are coplanar.Two lines (π₯ β π₯_1)/π_1 = (π¦ β π¦_1)/π_1 = (π§ β π§_1)/π_1 and (π₯ β π₯_2)/π_2 = (π¦ β π¦_2)/π_2 = (π§ β π§_2)/π_2 are coplanar if |β 8(π_πβ π_π&π_πβπ_π&π_πβπ_π@π_π&π_π&π_π@π_π&π_π&π_π )| = 0 (π β π + π )/(πΆ β πΉ) = (π β π)/πΆ = (π β π β π )/(πΆ + πΉ) (π₯ β (π β π))/(πΌ β πΏ) = (π¦ β π)/πΌ = (π§ β (π + π))/(πΌ + πΏ) Comparing (π₯ β π₯_1)/π_1 = (π¦ β π¦_1)/π_1 = (π§ β π§_1)/π_1 π₯_1 = π β d , π¦_1= π , π§_1= π + d & π_1=πΌβπΏ, π_1= πΌ, π_1= πΌ+πΏ (π β π + π)/(π· β πΈ) = (π β π)/π· = (π β π β π)/(π· + πΈ) (π₯ β (π β π))/(π½ β πΎ) = (π¦ β π)/π½ = (π§ β (π + π))/(π½ + πΎ) Comparing (π₯ β π₯_2)/π_2 = (π¦ β π¦_2)/π_2 = (π§ β π§_1)/π_2 π₯_2 = π β c , π¦_2= π , π§_2= π + c & π_2 = π½βπΎ, π_2 = π½, π_2 = π½ + πΎ Now, |β 8(π₯_2βπ₯_1&π¦_2βπ¦_1&π§_2βπ§_1@π_1&π_1&π_1@π_2&π_2&π_2 )| = |β 8(πβπβπ + π&πβπ&π+πβπβπ@πΌβπΏ&πΌ&πΌ+πΏ@π½βπΎ&π½&π½+πΎ)| Adding Column 3 to Column 1, = |β 8(πβπβπ + π+(π+πβπβπ)&πβπ&π+πβπβπ@πΌβπΏ+(πΌ+πΏ)&πΌ&πΌ+πΏ@π½βπΎ+(π½+πΎ)&π½&π½+πΎ)| = |β 8(2(πβπ)&πβπ&π+πβπβπ@2πΌ&πΌ&πΌ+πΏ@2π½&π½&π½+πΎ)| Taking 2 common from Column 1 = 2 |β 8(π β π&π β π&π + π β π β π@πΌ&πΌ&πΌ + πΏ@π½&π½&π½ +πΎ)| = 2 Γ 0 = 0 Therefore, the given two lines are coplanar. Since Columns 1 and 2 are same, The value of determinant is zero.