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If the shortest distance between the lines x-41= y+12= z-3 and x- 2= y+14= z-2-5 is 6 5, then the sum of all possible values of is :
(4) 8
a1=(4,-1,0), d1=(1,2,-3); a2=( ,-1,2), d2=(2,4,-5)
d1 d2= vmatrix i & j & k\ 1 & 2 & -3\ 2 & 4 & -5 vmatrix=2 i- j, | d1 d2|= 5
( a2- a1) ( d1 d2)=( -4,0,2) (2,-1,0)=2( -4)
SD = |2( -4)| 5= 6 5 | -4|=3
=7 or 1; sum =8
Answer: (4)
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