Slide10.JPG

Slide11.JPG
Slide12.JPG

Subscribe to our Youtube Channel - https://you.tube/teachoo

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

Transcript

Example 21 Show that the lines (๐‘ฅ + 3)/( โˆ’3) = (๐‘ฆ โˆ’ 1)/1 = (๐‘ง โˆ’ 5)/5 and (๐‘ฅ + 1)/( โˆ’1) = (๐‘ฆ โˆ’ 2)/2 = (๐‘ง โˆ’ 5)/5 are coplanar. Two lines (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 and (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_2)/๐‘_2 are coplanar if |โ– 8(๐’™_๐Ÿโˆ’๐’™_๐Ÿ&๐’š_๐Ÿโˆ’๐’š_๐Ÿ&๐’›_๐Ÿโˆ’๐’›_๐Ÿ@๐’‚_๐Ÿ&๐’ƒ_๐Ÿ&๐’„_๐Ÿ@๐’‚_๐Ÿ&๐’ƒ_๐Ÿ&๐’„_๐Ÿ )| = 0 Given, the two lines are Given, (๐’™ + ๐Ÿ‘)/( โˆ’ ๐Ÿ‘) = (๐’š โˆ’ ๐Ÿ)/๐Ÿ = (๐’› โˆ’ ๐Ÿ“)/๐Ÿ“ (๐‘ฅ โˆ’(โˆ’3))/( โˆ’ 3) = (๐‘ฆ โˆ’1)/1 = (๐‘งโˆ’ 5)/5 Comparing (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 ๐‘ฅ_1 = โˆ’3, ๐‘ฆ_1 = 1, ๐‘ง_1= 5 & ๐‘Ž_1 = โˆ’3, ๐‘_1 = 1, ๐‘_1= 5 Given, (๐’™ + ๐Ÿ)/( โˆ’ ๐Ÿ) = (๐’š โˆ’ ๐Ÿ)/๐Ÿ = (๐’› โˆ’ ๐Ÿ“)/๐Ÿ“ (๐‘ฅ โˆ’ (โˆ’1))/( โˆ’ 1) = (๐‘ฆ โˆ’ 2)/2 = (๐‘ง โˆ’ 5)/5 Comparing (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_2)/๐‘_2 ๐‘ฅ_2 = โˆ’1, ๐‘ฆ_2 = 2, ๐‘ง_2= 5 & ๐‘Ž_2 = โˆ’1, ๐‘_2 = 2, ๐‘_2= 5 Now, |โ– 8(๐‘ฅ_2โˆ’๐‘ฅ_1&๐‘ฆ_2โˆ’๐‘ฆ_1&๐‘ง_2โˆ’๐‘ง_1@๐‘Ž_1&๐‘_1&๐‘_1@๐‘Ž_2&๐‘_2&๐‘_2 )| `= |โ– 8( โˆ’1โˆ’(โˆ’3)&2โˆ’1&5โˆ’5@ โˆ’3&1&5@ โˆ’1&2&5)| = |โ– 8(2&1&0@โˆ’3&1&5@โˆ’1&2&5)| = 2[(1ร—5)โˆ’(2ร—5)] โˆ’ 1[(โˆ’3ร—5)โˆ’(โˆ’ 1ร—5)] + 0 [(โˆ’3ร—2)โˆ’(โˆ’1ร—1)] = 2[5โˆ’10]โˆ’ 1[โˆ’ 15โˆ’(โˆ’ 5)] + 0 = 2(โˆ’5) โˆ’1(โˆ’10) = โˆ’10 + 10 = 0 Therefore, the given two lines are coplanar.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.