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Question
Find the shortest distance between lines →r=6^i+2^j+2^k+λ(^i−2^j+2^k) and →r=−4^i−^k+μ(3^i−2^j−2^k)
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Solution
The given lines are
→r=6^i+2^j+2^k+λ(^i−2^j+2^k).......(1)
→r=−4^i−^k+μ(3^i−2^j−2^k).....(2)
It is known that the shortest distance between two lines, →r=→a1+λ→b1 and →r=→a2+λ→b2 is given by,
d=∣∣
∣
∣∣(→b1×→b2).(→a2−→a1)∣∣→b1×→b2∣∣∣∣
∣
∣∣........(3)
Comparing →r=→a1+λ→b1 and →r=→a2+λ→b2 to equations (1) and (2), we obtain
→a1=6^i+2^j+2^k
→b1=^i−2^j+2^k
→a2=−4^i−^k
→b2=3^i−2^j−2^k
⇒ →a2−→a1=(−4^i−^k)−(6^i+2^j+2^k)=10^i−2^j−3^k
⇒ ∣∣→b1×→b2∣∣=∣∣
∣
∣∣^i^j^k1−223−2−2∣∣
∣
∣∣=(4+4)^i−(−2−6)^j+(−2+6)^k=8^i+8^j+4^k
∴ ∣∣→b1×→b2∣∣=√(8)2+(8)2+(4)2=12
(→b1×→b2).(→a2−→a1)=((8^i+8^j+4^k).(−10^i−2^j−3^k)=−80−16−12=−108
Substituting all the values in equation (1), we obtain
d=∣∣∣−10812∣∣∣=9
Therefore, the shortest distance between the two given lines is 9 units.
→r=6^i+2^j+2^k+λ(^i−2^j+2^k).......(1)
→r=−4^i−^k+μ(3^i−2^j−2^k).....(2)
It is known that the shortest distance between two lines, →r=→a1+λ→b1 and →r=→a2+λ→b2 is given by,
d=∣∣ ∣ ∣∣(→b1×→b2).(→a2−→a1)∣∣→b1×→b2∣∣∣∣ ∣ ∣∣........(3)
Comparing →r=→a1+λ→b1 and →r=→a2+λ→b2 to equations (1) and (2), we obtain
→a1=6^i+2^j+2^k
→b1=^i−2^j+2^k
→a2=−4^i−^k
→b2=3^i−2^j−2^k
⇒ →a2−→a1=(−4^i−^k)−(6^i+2^j+2^k)=10^i−2^j−3^k
⇒ ∣∣→b1×→b2∣∣=∣∣ ∣ ∣∣^i^j^k1−223−2−2∣∣ ∣ ∣∣=(4+4)^i−(−2−6)^j+(−2+6)^k=8^i+8^j+4^k
∴ ∣∣→b1×→b2∣∣=√(8)2+(8)2+(4)2=12
(→b1×→b2).(→a2−→a1)=((8^i+8^j+4^k).(−10^i−2^j−3^k)=−80−16−12=−108
Substituting all the values in equation (1), we obtain
d=∣∣∣−10812∣∣∣=9
Therefore, the shortest distance between the two given lines is 9 units.
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