Miscellaneous

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Misc 3 If the lines (๐ฅ โ 1)/( โ 3) = (๐ฆ โ 2)/2๐ = (๐ง โ 3)/2 and (๐ฅ โ 1)/3๐ = (๐ฆ โ 1)/1 = (๐ง โ 6)/( โ 5) are perpendicular, find the value of k. Two lines (๐ฅ โ ๐ฅ1)/( ๐1) = (๐ฆ โ๐ฆ1)/๐1 = (๐ง โ ๐ง1)/๐1 and (๐ฅ โ ๐ฅ2)/( ๐2) = (๐ฆ โ ๐ฆ2)/๐2 = (๐ง โ ๐ง2)/๐2 are perpendicular to each other if ๐๐ ๐๐ + ๐๐ ๐๐ + ๐๐ ๐๐ = 0 (๐ โ ๐)/( โ ๐) = (๐ โ ๐)/( ๐๐) = (๐ โ ๐)/( ๐) Comparing with (๐ฅ โ ๐ฅ1)/( ๐1) = (๐ฆโ ๐ฆ1)/( ๐1) = (๐งโ ๐ง1)/( ๐1) x1 = 1, y1 = 2, z1 = 3 & ๐๐ = โ3, b1 = 2k c1 = 2 (๐ โ ๐)/( ๐๐) = (๐ โ ๐)/( ๐) = (๐ โ ๐)/( โ ๐) Comparing with (๐ฅ โ ๐ฅ2)/( ๐2) = (๐ฆ โ ๐ฆ2)/( ๐2) = (๐ง โ ๐ง2)/( ๐2), ๐ฅ2 = 1, y2 = 1, z2 = 6 & ๐๐ = 3k, b2 = 1, c2 = โ5, Since the two lines are perpendicular, ๐1 ๐2 + ๐1 ๐2 + c1๐2 = 0 (โ3 ร 3k) + (2k ร 1) + (2 ร โ 5) = 0 โ 9k + 2k โ 10 = 0 โ 7k = 10 k = (โ๐๐)/๐ Therefore, k = (โ10)/7