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Let `l_(1) and l_(2)` be the two skew lines. If P, Q are two distinct points on `l_(1)` and R, S are two distinct points on `l_(2)`, then prove that PR cannot be parallel to QS. |
Answer» Let equation of the line `l_(1)` be `vecr=veca+lamdavecb` and equation of the line `l_(2)` be `vecr=vecc+muvecd`, where `veca-vecc, vecb and vecd` are non-coplanar. Let the position vectors of points P and Q be `veca+lamda_(1)vecb and vecaa+lamda_(2)vecb`, respectively. Let the position vectors of points R and S be `vecc+mu_(1)vecd and vecc+mu_(1)vecd`, respectively. Then the lines PR and QS are parallel if and only if `vecc-veca+mu_(1)vecd-lamda_(1)vecb=k(vecc-veca+mu_(2)vecd-lamda_(2)vecb)` i.e., `" "(1-k)(vecc-veca)+(mu_(1)-kmu_(2))vecd - (lamda_(1)-klamda_(2))vecb=veco` `therefore" "1-k=0, mu_(1)-kmu_(2)=0, lamda_(1)-klamda_(2)=0` i.e., `" "mu_(1)=mu_(2) and lamda_(1)=lamda_(2)` which is not possible Therefore, PR can not be parallel to QS. |
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