Find the shortest distance between the 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)
Find the shortest distance between the 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)
Given lines are r=6^i+2^j+2^k+λ(^i−2^j+2^k) . . .(i)
r=−4^i−^k+μ(3^i−2^j−2^k) . . .(ii)
It is known that the shortest distance between two lines r=a1+λb1 and r=a2+μb2 is given by d=|(b1×b2).(a2−a1)||b1×b2| . . .(iii)
Comparing r=a1+λb1 and r=a2+μb2 to Eqs. (i) and (ii), we obtain
a1=6^i+2^j+2^k, a2=−4^i−^k, b1=^i−2^j+2^k,b2=3^i−2^j−2^kNow, a2−a1=(−4^i−^k)−(6^i+2^j+2^k)=−10^i−2^j−3^k
Also, b1×b2=∣∣
∣
∣∣^i^j^k1−223−2−2∣∣
∣
∣∣=^i(4+4)−^j(−2−6)+^k(−2+6)=8^i+8^j+4^[k]
∴ |b1×b2|=√82+82+42=√64+64+16=√144=12Now, (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 Eq. (iii), we obtain
d=|(b1×b2).(a2−a1)||b1×b2|=10812=9 units
Given lines are r=6^i+2^j+2^k+λ(^i−2^j+2^k) . . .(i)
r=−4^i−^k+μ(3^i−2^j−2^k) . . .(ii)
It is known that the shortest distance between two lines r=a1+λb1 and r=a2+μb2 is given by d=|(b1×b2).(a2−a1)||b1×b2| . . .(iii)
Comparing r=a1+λb1 and r=a2+μb2 to Eqs. (i) and (ii), we obtain
a1=6^i+2^j+2^k, a2=−4^i−^k, b1=^i−2^j+2^k,b2=3^i−2^j−2^kNow, a2−a1=(−4^i−^k)−(6^i+2^j+2^k)=−10^i−2^j−3^k
Also, b1×b2=∣∣
∣
∣∣^i^j^k1−223−2−2∣∣
∣
∣∣=^i(4+4)−^j(−2−6)+^k(−2+6)=8^i+8^j+4^[k]
∴ |b1×b2|=√82+82+42=√64+64+16=√144=12Now, (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 Eq. (iii), we obtain
d=|(b1×b2).(a2−a1)||b1×b2|=10812=9 units
.