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Vertices of a parallelogram taken in ord...

Vertices of a parallelogram taken in order are A, ( 2,-1,4) , B (1,0,-1) , C ( 1,2,3) and D.
The distance between the parallel lines AB and CD is

A

`sqrt6`

B

`3sqrt(6//5)`

C

`2sqrt2`

D

3

Text Solution

Verified by Experts

The correct Answer is:
c
Let point D be `(a_(1),a_(2),a_(3))`
`a_(1)+1=3or a_(1)=2`
`a_(2) +0= 1 l or a_(2)=1`
`a_(3)-1 = 7 or a_(3)=8`
`a_(3)-1=7 or a_(3)=8`
`vecd=||vec((AB))xxvec((AD))|/|vec(AB)||`
`vec(AB)= -hati+hatj-5hatk`
`vec(AD)=0hati+2hatj=4hatk`
`vec(AB)xx vec(AD)=|{:(hati,hatj,hatk),(-1,1,-5),(0,2,4):}|`
`14hati+4hatj-2hatk`
`2(7hati+2hatj-hatk)`
`Rightarrow d=2sqrt2`
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