InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Determine whether the following pair of lines intersect or not. i. `vecr=hati-hatj+lamda(2hati+hatk), vecr=2hati-hatj+mu(hati+hatj-hatk)` |
|
Answer» Here `vec(a_(1))=veci-vecj, vec(a_(2))=2hati -hatj, vec(b_(1))=2hati+hatk and vec(b_(2))=hati+hatj-hatk` Now `" "[vec(a_(2))-vec(a_(1))vec(b_(1))vec(b_(2))]=|{:(2-1, ,-1+1,,0),(2,,0,,1),(1,,1,,-1):}|` `" "=|{:(1,,0,,0),(2,,0,,1),(1,,1,,-1):}|` `" "=-1 ne 0` Thus, the two given lines do not intersect. ii. Here `vec(a_(1))=hati+hatj-hatk, vec(a_(2))=4hati-hatk, vec(b_(1))=3hati-hatj and vec(b_(2))=2hati+ 3hatk` `rArr" "[vec(a_(2))-vec(a_(1))vec(b_(1))vec(b_(2))]=|{:(4-1,,0-1,,-1+1),(3,,-1,,0),(2,,0,,3):}|` `" "=|{:(3,,-1,,0),(3,,-1,,0),(2,,0,,3):}|=0` Thus, the two given lines intersect. Let us obtain the point of intersection of the two given lines. For some values of `lamda` and `mu`, the two values of `vecr` must coincide. Thus, `" "hati+hatj-hatk+ lamda(3hati-hatj)=4hati-hatk+mu(2hati+3hatk)` or `" "(3+2mu-3lamda)hati+(lamda-1)hatj+3muhatk=0` or `" "3+2mu-3lamda=0, lamda-1=0, 3mu=0` Solving, we obtain `lamda=1 and mu=0` Therefore, the point of intersection is `vecr=4hati-hatk`( by putting `mu=0` in the second equation). |
|