已知空间四点A(1,-2,1)B(2,-1,2)C(3,2,-1)D(1,1,-1),求点E,使向量DE⊥向量AB,向量DE⊥向量AC,

且绝对值DE向量=根号14
2025-01-06 02:22:25
推荐回答(1个)
回答1:

AB=OB-OA=(2,-1,2)-(1,-2,1)=(1,1,1),AC=OC-OA=(3,2,-1)-(1,-2,1)=(2,4,-2)
设E点为(x,y,z),则:DE=OE-OD=(x,y,z)-(1,1,-1)=(x-1,y-1,z+1)
而:AB×AC=(1,1,1)×(2,4,-2)=-6i+4j+2k=(-6,4,2),即:|AB×AC|=sqrt(56)=2sqrt(14)
故所求向量DE的模值是|AB×AC|的1/2,即:DE=AB×AC/2或-AB×AC/2=AC×AB/2
即:DE=(-3,2,1)或DE=(3,-2,-1),即:(x-1,y-1,z+1)=(-3,2,1)或(3,-2,-1)
故:E点为(-2,3,0)或(4,-1,-2)