Trong không gian Oxyz, cho điểm M(3; -4; -5) và các đường thẳng

Gọi A−4−5a;4+2a;2+3a∈d1,B1−b;2+3b;−5−2b∈d2.

Vì M,A,B∈d nên chúng thẳng hàng ⇒MA→,MB→ cùng phương.

Ta có: MA→=5a+7;−2a−8;−3a−7MB→=b+2;−3b−6;2b

⇒5a+7b+2=−2a−8−3b−6=−3a−72b

⇔15ab+30a+21b+42=2ab+8b+4a+1610ab+14b=−3ab−6a−7b−14

⇔13ab+26a+13b+26=013ab+6a+21b+14=0

⇔20a−8b+12=0ab+2a+b+2=0

⇔5a−2b+3=0ab+2a+b+2=0

⇔b=5a+32a.5a+32+2a+5a+32+2=0

⇔b=5a+325a2+3a+4a+5a+3+4=0

⇔b=5a+325a2+12a+7=0

⇔a=−1,b=−1a=−75,b=−2

⇒A1;2;−1,B2;−1;−3A3;65;−115,B3;−4;−1

 

TH1: A1;2;−1,B2;−1;−3⇒OA→1;2;−1,OB→2;−1;−3

⇒SOAB=12OA→,OB→=1232+62+02=352

TH2: A3;65;−115,B3;−4;−1⇒OA→3;65;−115,OB→3;−4;−1

⇒SOAB=12OA→,OB→=12.102+3,62+15,62=890810.

Chọn B.