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Question
Find the shortest distance between lines: x+1−10=y+3−1=z−41 and x+10−1=y+1−3=z−14
Find the shortest distance between lines: x+1−10=y+3−1=z−41 and x+10−1=y+1−3=z−14
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Solution
The correct option is C 0
Given lines
l1:x+1−10=y+3−1=z−41
l2:x+10−1=y+1−3=z−14
position vector of line l1→a=−^i−3^j+4^k
position vector of line l2→c=−10^i−^j+^k
normal vector of line l1→n1=−10^i−^j+^k
normal vector of line l2→n2=−^i−3^j+4^k
so line are skews line SD=∣∣∣(→c−→a)⋅(→n1×→n2)|(→n1×→n2)|∣∣∣
→c−→a=−10^i−^j+^k+^i+3^j−4^k
→c−→a=−9^i+2^j−3^k
→n1×→n2=∣∣
∣
∣∣^i^j^k−10−11−1−34∣∣
∣
∣∣
→n1×→n2=^i(−4+3)−^j(−40+1)+^k(30−1)
→n1×→n2=−^i+39^j+29^k
|→n1×→n2|=√392+292+(−1)2
|→n1×→n2|=√2363
putting →c−→a,→n1×→n2,|→n1×→n2| in formula
SD=∣∣
∣∣(−9^i+2^j−3^k)⋅(−^i+39^j+29^k√2363∣∣
∣∣
SD=∣∣∣9+78−87√2363∣∣∣
SD=∣∣∣0√2363∣∣∣
SD=0
Given lines
l1:x+1−10=y+3−1=z−41
l2:x+10−1=y+1−3=z−14
position vector of line l1
→a=−^i−3^j+4^k
position vector of line l2
→c=−10^i−^j+^k
normal vector of line l1
→n1=−10^i−^j+^k
normal vector of line l2
→n2=−^i−3^j+4^k
so line are skews line
SD=∣∣∣(→c−→a)⋅(→n1×→n2)|(→n1×→n2)|∣∣∣
→c−→a=−10^i−^j+^k+^i+3^j−4^k
→c−→a=−9^i+2^j−3^k
→n1×→n2=∣∣
∣
∣∣^i^j^k−10−11−1−34∣∣
∣
∣∣
→n1×→n2=^i(−4+3)−^j(−40+1)+^k(30−1)
→n1×→n2=−^i+39^j+29^k
|→n1×→n2|=√392+292+(−1)2
|→n1×→n2|=√2363
putting →c−→a,→n1×→n2,|→n1×→n2| in formula
SD=∣∣
∣∣(−9^i+2^j−3^k)⋅(−^i+39^j+29^k√2363∣∣
∣∣
SD=∣∣∣9+78−87√2363∣∣∣
SD=∣∣∣0√2363∣∣∣
SD=0
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