Wehavethegivenequation→r=(6ˆi+2ˆj+2ˆk)+λ(ˆi−2ˆj+2ˆk)comparingwith→r=→a1+λ→b1→a1=(6ˆi+2ˆj+2ˆk)&→b1=(ˆi−2ˆj+2ˆk)similarly→r=(−4ˆi−ˆk)+μ(3ˆi−2ˆj−2ˆk)comparingwith→r=→a2+μ→b2→a2=(−4ˆi−ˆk)&→b2=(3ˆi−2ˆj−2ˆk)Now,(→a2−→a2)=(−4ˆi−ˆk)−(6ˆi+2ˆj+2ˆk)=−10ˆi−2ˆj−3ˆk(→b1×→b2)=∣∣
∣
∣∣ˆiˆjˆk1−223−2−2∣∣
∣
∣∣=ˆi[(−2×−2)−(−2×2)]−ˆj[(1×−2)−(3×2)]+ˆk[(1×−2)−(3×−2)]=ˆi[4+4]+ˆj[−2−6]+ˆk[−2+6]=8ˆi+8ˆj+4ˆkNowMagnitudeof→b1×→b2=√82+82+42=√64+64+16=√144=12Also,(→b1×→b2).(→a2−→a2)=(−10ˆi−2ˆj−3ˆk)(8ˆi+8ˆj+4ˆk)=(8×−10)+(8×−2)+(4×−3)=−80−16−12=−108Shortestdistance=∣∣
∣∣(→b1×→b2).(→a2−→a2)(→b1×→b2)∣∣
∣∣=∣∣−10812∣∣=|−9|=9Hence,theshortestdistancebwtweenthegivenlineis9.