1
Question
Find the shortest distance between lines: x−61=y−2−2=z−22 and x+43=y−2=z+1−2
Find the shortest distance between lines: x−61=y−2−2=z−22 and x+43=y−2=z+1−2
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Solution
The correct option is C 9
Given linesl1:x−61=y−2−2=z−22
l2:x+43=y−2=z+1−2
position vector of line l1→a=6^i+2^j+2^k
position vector of line l2→c=−4^i−^k
normal vector of line l1→n1=^i−2^j+2^k
normal vector of line l2→n2=3^i−2^j−2^k
so line are skews line SD=∣∣∣(→c−→a)⋅(→n1×→n2)|(→n1×→n2)|∣∣∣
→c−→a=−4^i−^k−6^i−2^j−2^k
→c−→a=−10^i−2^j−3^k
→n1×→n2=∣∣
∣
∣∣^i^j^k1−223−2−2∣∣
∣
∣∣
→n1×→n2=^i(4+4)−^j(−2−6)+^k(−2+6)
→n1×→n2=8^i+8^j+4^k
|→n1×→n2|=√82+82+(4)2
|→n1×→n2|=√144=12
putting →c−→a,→n1×→n2,|→n1×→n2| in formula
SD=∣∣
∣∣(−10^i−2^j−3^k)⋅(8^i+8^j+4^k12∣∣
∣∣
SD=∣∣∣−80−16−1212∣∣∣
SD=∣∣∣−10812∣∣∣
SD=10812
SD=9
Given lines
l1:x−61=y−2−2=z−22
l2:x+43=y−2=z+1−2
position vector of line l1
→a=6^i+2^j+2^k
position vector of line l2
→c=−4^i−^k
normal vector of line l1
→n1=^i−2^j+2^k
normal vector of line l2
→n2=3^i−2^j−2^k
so line are skews line
SD=∣∣∣(→c−→a)⋅(→n1×→n2)|(→n1×→n2)|∣∣∣
→c−→a=−4^i−^k−6^i−2^j−2^k
→c−→a=−10^i−2^j−3^k
→n1×→n2=∣∣
∣
∣∣^i^j^k1−223−2−2∣∣
∣
∣∣
→n1×→n2=^i(4+4)−^j(−2−6)+^k(−2+6)
→n1×→n2=8^i+8^j+4^k
|→n1×→n2|=√82+82+(4)2
|→n1×→n2|=√144=12
putting →c−→a,→n1×→n2,|→n1×→n2| in formula
SD=∣∣
∣∣(−10^i−2^j−3^k)⋅(8^i+8^j+4^k12∣∣
∣∣
SD=∣∣∣−80−16−1212∣∣∣
SD=∣∣∣−10812∣∣∣
SD=10812
SD=9
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