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Question

Find the shortest distance between the following pairs of parallel lines whose equations are:
(i) r=i^+2j^+3k^+λi^-j^+k^ and r=2i^-j^-k^+μ-i^+j^-k^

(ii) r=i^+j^+λ2i^-j^+k^ and r=2i^+j^-k^+μ4i^-2j^+2k^

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Solution

(i) The vector equations of the given lines are

r=i^+2j^+3k^+λi^-j^+k^ ...(1)r=2i^-j^-k^+μ-i^+j^-k^ =2i^-j^-k^-μi^-j^+k^ ...(2)

These two lines pass through the points having position vectors a1=i^+2j^+3k^ and a2=2i^-j^-k^ and are parallel to the vector b=i^-j^+k^.

Now,
a2-a1=i^-3j^-4k^
and
a2-a1×b=i^-3j^-4k^×i^-j^+k^ =i^j^k^1-3-41-11 =-7i^-5j^+2k^a2-a1×b=-72+-52+22 =49+25+4 =78

The shortest distance between the two lines is given by

a2-a1×bb=783=26

(ii) r=i^+j^+λ2i^-j^+k^ and r=2i^+j^-k^+μ4i^-2j^+2k^ or r=2i^+j^-k^+2μ2i^-j^+k^

These two lines pass through the points having position vectors a1=i^+j^ and a2=2i^+j^-k^ and are parallel to the vector b=2i^-j^+k^.

Now,
a2-a1=i^-k^
and
a2-a1×b=i^-k^×2i^-j^+k^ =i^j^k^10-12-11 =-i^-3j^-k^a2-a1×b=-12+-32+-12 =1+9+1 =11

The shortest distance between the two lines is given by

a2-a1×bb=116

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