Question

已知向量OA=3I-4J,OB=6I-3J,向量OC=(5-M)I-(3+M)J,其中I,J分别是

Answer 1

若A,B,C能构成三角形,

则AB,AC,BC相互不平行

AB=6I-3J-3I+4J=3I+J

AC=3I-4J-(5-M)I+(3+M)J=(-2+M)I+(-1+M)J

BC=6I-3J-(5-M)I+(3+M)J=(1+M)I+MJ

即(-2+M)/3≠(-1+M)/1 解得M≠1/2

(1+M)/3≠M/1 解得M≠1/2

(1+M)/(-2+M)≠M/(-1+M) 解得M≠1/2

综上：实数M应满足M≠1/2

2.AB⊥AC

则3(-2+M)+1*(-1+M)=0

解得M=7/4