Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Example 21 Show that the lines (π₯ + 3)/( β3) = (π¦ β 1)/1 = (π§ β 5)/5 and (π₯ + 1)/( β1) = (π¦ β 2)/2 = (π§ β 5)/5 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_πβπ_π&π_πβπ_π&π_πβπ_π@π_π&π_π&π_π@π_π&π_π&π_π )| = 0 Given, the two lines are Given, (π + π)/( β π) = (π β π)/π = (π β π)/π (π₯ β(β3))/( β 3) = (π¦ β1)/1 = (π§β 5)/5 Comparing (π₯ β π₯_1)/π_1 = (π¦ β π¦_1)/π_1 = (π§ β π§_1)/π_1 π₯_1 = β3, π¦_1 = 1, π§_1= 5 & π_1 = β3, π_1 = 1, π_1= 5 Given, (π + π)/( β π) = (π β π)/π = (π β π)/π (π₯ β (β1))/( β 1) = (π¦ β 2)/2 = (π§ β 5)/5 Comparing (π₯ β π₯_2)/π_2 = (π¦ β π¦_2)/π_2 = (π§ β π§_2)/π_2 π₯_2 = β1, π¦_2 = 2, π§_2= 5 & π_2 = β1, π_2 = 2, π_2= 5 Now, |β 8(π₯_2βπ₯_1&π¦_2βπ¦_1&π§_2βπ§[email protected]π_1&π_1&π[email protected]π_2&π_2&π_2 )| `= |β 8( β1β(β3)&2β1&5β[email protected] β3&1&[email protected] β1&2&5)| = |β 8(2&1&[email protected]β3&1&[email protected]β1&2&5)| = 2[(1Γ5)β(2Γ5)] β 1[(β3Γ5)β(β 1Γ5)] + 0 [(β3Γ2)β(β1Γ1)] = 2[5β10]β 1[β 15β(β 5)] + 0 = 2(β5) β1(β10) = β10 + 10 = 0 Therefore, the given two lines are coplanar.