Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Question 11 Show that the lines (๐ฅ + 3)/( โ3) = (๐ฆ โ 1)/1 = (๐ง โ 5)/5 and (๐ฅ + 1)/( โ1) = (๐ฆ โ 2)/2 = (๐ง โ 5)/5 are coplanar. Two lines (๐ฅ โ ๐ฅ_1)/๐_1 = (๐ฆ โ ๐ฆ_1)/๐_1 = (๐ง โ ๐ง_1)/๐_1 and (๐ฅ โ ๐ฅ_2)/๐_2 = (๐ฆ โ ๐ฆ_2)/๐_2 = (๐ง โ ๐ง_2)/๐_2 are coplanar if |โ 8(๐_๐โ๐_๐&๐_๐โ๐_๐&๐_๐โ๐_๐@๐_๐&๐_๐&๐_๐@๐_๐&๐_๐&๐_๐ )| = 0 Given, the two lines are Given, (๐ + ๐)/( โ ๐) = (๐ โ ๐)/๐ = (๐ โ ๐)/๐ (๐ฅ โ(โ3))/( โ 3) = (๐ฆ โ1)/1 = (๐งโ 5)/5 Comparing (๐ฅ โ ๐ฅ_1)/๐_1 = (๐ฆ โ ๐ฆ_1)/๐_1 = (๐ง โ ๐ง_1)/๐_1 ๐ฅ_1 = โ3, ๐ฆ_1 = 1, ๐ง_1= 5 & ๐_1 = โ3, ๐_1 = 1, ๐_1= 5 Given, (๐ + ๐)/( โ ๐) = (๐ โ ๐)/๐ = (๐ โ ๐)/๐ (๐ฅ โ (โ1))/( โ 1) = (๐ฆ โ 2)/2 = (๐ง โ 5)/5 Comparing (๐ฅ โ ๐ฅ_2)/๐_2 = (๐ฆ โ ๐ฆ_2)/๐_2 = (๐ง โ ๐ง_2)/๐_2 ๐ฅ_2 = โ1, ๐ฆ_2 = 2, ๐ง_2= 5 & ๐_2 = โ1, ๐_2 = 2, ๐_2= 5 Now, |โ 8(๐ฅ_2โ๐ฅ_1&๐ฆ_2โ๐ฆ_1&๐ง_2โ๐ง_1@๐_1&๐_1&๐_1@๐_2&๐_2&๐_2 )| `= |โ 8( โ1โ(โ3)&2โ1&5โ5@ โ3&1&5@ โ1&2&5)| = |โ 8(2&1&0@โ3&1&5@โ1&2&5)| = 2[(1ร5)โ(2ร5)] โ 1[(โ3ร5)โ(โ 1ร5)] + 0 [(โ3ร2)โ(โ1ร1)] = 2[5โ10]โ 1[โ 15โ(โ 5)] + 0 = 2(โ5) โ1(โ10) = โ10 + 10 = 0 Therefore, the given two lines are coplanar.