Find the angle between the lines `vecr=hati-hatj+hatk+lambda(2hati-2hatj+hatk)` and `vecr=2hati-hatj+2hatk+mu(hati+hatj+2hatk)`.

Explanation

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).