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Find the shortest distance between the skew lines r=(6ˆi+2ˆj+2ˆk)+λ(ˆi2ˆj+2ˆk) and r=(4ˆiˆk)+μ(3ˆi2ˆj2ˆk)

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Solution

Wehavethegivenequationr=(6ˆi+2ˆj+2ˆk)+λ(ˆi2ˆj+2ˆk)comparingwithr=a1+λb1a1=(6ˆi+2ˆj+2ˆk)&b1=(ˆi2ˆj+2ˆk)similarlyr=(4ˆiˆk)+μ(3ˆi2ˆj2ˆk)comparingwithr=a2+μb2a2=(4ˆiˆk)&b2=(3ˆi2ˆj2ˆk)Now,(a2a2)=(4ˆiˆk)(6ˆi+2ˆj+2ˆk)=10ˆi2ˆj3ˆk(b1×b2)=∣ ∣ ∣ˆiˆjˆk122322∣ ∣ ∣=ˆi[(2×2)(2×2)]ˆj[(1×2)(3×2)]+ˆk[(1×2)(3×2)]=ˆi[4+4]+ˆj[26]+ˆk[2+6]=8ˆi+8ˆj+4ˆkNowMagnitudeofb1×b2=82+82+42=64+64+16=144=12Also,(b1×b2).(a2a2)=(10ˆi2ˆj3ˆk)(8ˆi+8ˆj+4ˆk)=(8×10)+(8×2)+(4×3)=801612=108Shortestdistance=∣ ∣(b1×b2).(a2a2)(b1×b2)∣ ∣=10812=|9|=9Hence,theshortestdistancebwtweenthegivenlineis9.

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