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Question
If A=(1,−2,−1),B=(4,0,−3);C=(1,2,−1) and D=(2,−4,−5) , then the distance between AB and CD is
If A=(1,−2,−1),B=(4,0,−3);C=(1,2,−1) and D=(2,−4,−5) , then the distance between AB and CD is
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Solution
The correct option is D 43
AB=→r1+b1¯AB=(^i−2^j−^k)+t(4^i−3^k−^i+2^j+^k)¯AB=(^i−2^j−^k)+t(3^i+2^j−2^k)→CD=→r2+t→b2+(^i+2^j−^k)+t(2^i−4^j−5^k)−(^i+2^j^k)→CD=(^i+2^j−^k)+t(^i−6^j−4^4)→n=¯b1ׯb2=∣∣
∣
∣∣^i^j^k32−21−6−4∣∣
∣
∣∣=−20^i−^j(−12+2)^k(−18−2)→n=−20^i+10^j−20^kd=→n.(→r1−→r2)|→n|=(−20^i+10^j−20^k)(^i−2^j−^k−^i−2^j+^k)√(20)2+(10)2+(20)2
d=|(−20^i+10^j−20^k).(−4g)|30d=4030=d=43
AB=→r1+b1
¯AB=(^i−2^j−^k)+t(4^i−3^k−^i+2^j+^k)
¯AB=(^i−2^j−^k)+t(3^i+2^j−2^k)
→CD=→r2+t→b2
+(^i+2^j−^k)+t(2^i−4^j−5^k)−(^i+2^j^k)
→CD=(^i+2^j−^k)+t(^i−6^j−4^4)
→n=¯b1ׯb2=∣∣
∣
∣∣^i^j^k32−21−6−4∣∣
∣
∣∣
=−20^i−^j(−12+2)^k(−18−2)
→n=−20^i+10^j−20^k
d=→n.(→r1−→r2)|→n|=(−20^i+10^j−20^k)(^i−2^j−^k−^i−2^j+^k)√(20)2+(10)2+(20)2
d=|(−20^i+10^j−20^k).(−4g)|30
d=4030=d=43
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