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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Example 29 Show that the lines (π‘₯ βˆ’ π‘Ž + 𝑑)/(𝛼 βˆ’ 𝛿) = (𝑦 βˆ’ π‘Ž)/𝛼 = (𝑧 βˆ’ π‘Ž βˆ’ 𝑑)/(𝛼 + 𝛿) nd (π‘₯ βˆ’ 𝑏 + 𝑐)/(𝛽 βˆ’ 𝛾) = (𝑦 βˆ’ 𝑏)/𝛽 = (𝑧 βˆ’ 𝑏 βˆ’ 𝑐)/(𝛽 + 𝛾) are coplanar.Two lines (π‘₯ βˆ’ π‘₯_1)/π‘Ž_1 = (𝑦 βˆ’ 𝑦_1)/𝑏_1 = (𝑧 βˆ’ 𝑧_1)/𝑐_1 and (π‘₯ βˆ’ π‘₯_2)/π‘Ž_2 = (𝑦 βˆ’ 𝑦_2)/𝑏_2 = (𝑧 βˆ’ 𝑧_2)/𝑐_2 are coplanar if |β– 8(𝒙_πŸβˆ’ 𝒙_𝟏&π’š_πŸβˆ’π’š_𝟏&𝒛_πŸβˆ’π’›_𝟏@𝒂_𝟏&𝒃_𝟏&𝒄_𝟏@𝒂_𝟐&𝒃_𝟐&𝒄_𝟐 )| = 0 (𝒙 βˆ’ 𝒂 + 𝒅)/(𝜢 βˆ’ 𝜹) = (π’š βˆ’ 𝒂)/𝜢 = (𝒛 βˆ’ 𝒂 βˆ’ 𝒅)/(𝜢 + 𝜹) (π‘₯ βˆ’ (π‘Ž βˆ’ 𝑑))/(𝛼 βˆ’ 𝛿) = (𝑦 βˆ’ π‘Ž)/𝛼 = (𝑧 βˆ’ (π‘Ž + 𝑑))/(𝛼 + 𝛿) Comparing (π‘₯ βˆ’ π‘₯_1)/π‘Ž_1 = (𝑦 βˆ’ 𝑦_1)/𝑏_1 = (𝑧 βˆ’ 𝑧_1)/𝑐_1 π‘₯_1 = π‘Ž βˆ’ d , 𝑦_1= π‘Ž , 𝑧_1= π‘Ž + d & π‘Ž_1=π›Όβˆ’π›Ώ, 𝑏_1= 𝛼, 𝑐_1= 𝛼+𝛿 (𝒙 βˆ’ 𝒃 + 𝒄)/(𝜷 βˆ’ 𝜸) = (π’š βˆ’ 𝒃)/𝜷 = (𝒛 βˆ’ 𝒃 βˆ’ 𝒄)/(𝜷 + 𝜸) (π‘₯ βˆ’ (𝑏 βˆ’ 𝑐))/(𝛽 βˆ’ 𝛾) = (𝑦 βˆ’ 𝑏)/𝛽 = (𝑧 βˆ’ (𝑏 + 𝑐))/(𝛽 + 𝛾) Comparing (π‘₯ βˆ’ π‘₯_2)/π‘Ž_2 = (𝑦 βˆ’ 𝑦_2)/𝑏_2 = (𝑧 βˆ’ 𝑧_1)/𝑐_2 π‘₯_2 = 𝑏 βˆ’ c , 𝑦_2= 𝑏 , 𝑧_2= 𝑏 + c & π‘Ž_2 = π›½βˆ’π›Ύ, 𝑏_2 = 𝛽, 𝑐_2 = 𝛽 + 𝛾 Now, |β– 8(π‘₯_2βˆ’π‘₯_1&𝑦_2βˆ’π‘¦_1&𝑧_2βˆ’π‘§_1@π‘Ž_1&𝑏_1&𝑐_1@π‘Ž_2&𝑏_2&𝑐_2 )| = |β– 8(π‘βˆ’π‘βˆ’π‘Ž + 𝑑&π‘βˆ’π‘Ž&𝑏+π‘βˆ’π‘Žβˆ’π‘‘@π›Όβˆ’π›Ώ&𝛼&𝛼+𝛿@π›½βˆ’π›Ύ&𝛽&𝛽+𝛾)| Adding Column 3 to Column 1, = |β– 8(π‘βˆ’π‘βˆ’π‘Ž + 𝑑+(𝑏+π‘βˆ’π‘Žβˆ’π‘‘)&π‘βˆ’π‘Ž&𝑏+π‘βˆ’π‘Žβˆ’π‘‘@π›Όβˆ’π›Ώ+(𝛼+𝛿)&𝛼&𝛼+𝛿@π›½βˆ’π›Ύ+(𝛽+𝛾)&𝛽&𝛽+𝛾)| = |β– 8(2(π‘βˆ’π‘Ž)&π‘βˆ’π‘Ž&𝑏+π‘βˆ’π‘Žβˆ’π‘‘@2𝛼&𝛼&𝛼+𝛿@2𝛽&𝛽&𝛽+𝛾)| Taking 2 common from Column 1 = 2 |β– 8(𝑏 βˆ’ π‘Ž&𝑏 βˆ’ π‘Ž&𝑏 + 𝑐 βˆ’ π‘Ž βˆ’ 𝑑@𝛼&𝛼&𝛼 + 𝛿@𝛽&𝛽&𝛽 +𝛾)| = 2 Γ— 0 = 0 Therefore, the given two lines are coplanar. Since Columns 1 and 2 are same, The value of determinant is zero.

About the Author

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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.