4
Question
The distance between lines 3xā26=4yā312=4xā38 and 2xā12=3yā22=4zā3ā8
The distance between lines 3xā26=4yā312=4xā38 and 2xā12=3yā22=4zā3ā8
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Solution
The correct option is C 0.02
Given lines 3x−26=4y−312=4z−38x−232=y−343=z−342------(1)2x−12=3y−22=4z−3−8x−121=y−2323=z−34−2------(2)on comparing both lines with general cartesian form normal vector of line eq (1) →b=2^i+3^j+2^knormal vector of line eq (2) →d=^i+23^j−2^kposition vector of line eq (1)→a=23^i+34^j+34^kposition vector of line (2)→c=12^i+23^j+34^k→c−→a=12^i+23^j+34^k−(23^i+34^j+34^k)→c−→a=12^i+23^j+34^k−23^i−34^j−34^k→c−→a=−16^i−112^j→b×→d=∣∣
∣
∣
∣∣^i^j^k234123−2∣∣
∣
∣
∣∣→b×→d=^i(−6−83)−^j(−4−4)+^k(43−3)→b×→d=263^i+8^j−53^kformula to find distance between skews lines SD=∣∣(→c−→a)⋅(→b×→d)∣∣∣∣→b×→d∣∣SD=∣∣∣(−16^i−112^j)⋅(263^i+8^j−53^k)∣∣∣√(263)2+82+(−53)2SD=∣∣∣−2618−812+0∣∣∣√6769+16+259SD=7636√8459SD=0.02
Given lines
3x−26=4y−312=4z−38
x−232=y−343=z−342------(1)
2x−12=3y−22=4z−3−8
x−121=y−2323=z−34−2------(2)
on comparing both lines with general cartesian form
normal vector of line eq (1)
→b=2^i+3^j+2^k
normal vector of line eq (2)
→d=^i+23^j−2^k
position vector of line eq (1)
→a=23^i+34^j+34^k
position vector of line (2)
→c=12^i+23^j+34^k
→c−→a=12^i+23^j+34^k−(23^i+34^j+34^k)
→c−→a=12^i+23^j+34^k−23^i−34^j−34^k
→c−→a=−16^i−112^j
→b×→d=∣∣
∣
∣
∣∣^i^j^k234123−2∣∣
∣
∣
∣∣
→b×→d=^i(−6−83)−^j(−4−4)+^k(43−3)
→b×→d=263^i+8^j−53^k
formula to find distance between skews lines
SD=∣∣(→c−→a)⋅(→b×→d)∣∣∣∣→b×→d∣∣
SD=∣∣∣(−16^i−112^j)⋅(263^i+8^j−53^k)∣∣∣√(263)2+82+(−53)2
SD=∣∣∣−2618−812+0∣∣∣√6769+16+259
SD=7636√8459
SD=0.02
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