
数学向量,需要过程
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约定:用AB'表示“向量AB”, 用AC'表示“向量AC”,...
选:B
|AB'|=1 (1)
|EF'|=|(1/2)(AB'+AC')-(1/2)AD'|=√2
|AB'-(AD'-AC')|=2√2 (2)
|CD'|=|AD'-AC'|=√5
即 |AD'-AC'|=√5 (3)
AD'·BC'=AD'·(AC'-AB')=15
即AD'·AC'-AD'·AB'=15 (4)
由(2) 得
|AB'|²-2AB'·(AD'-AC')+|AD'-AC'|²=8
(1)(3)代入
1²-2AB'·(AD'-AC')+(√5)²=8
AB'·AC'-AB'·AD'=1 (5)
(4)-(5):
AD'·AC'-AB'·AC'=14
AC'·(AD'-AB')=AC'·BD'=14
所以 AC'·BD'=14
希望能帮到你!
选:B
|AB'|=1 (1)
|EF'|=|(1/2)(AB'+AC')-(1/2)AD'|=√2
|AB'-(AD'-AC')|=2√2 (2)
|CD'|=|AD'-AC'|=√5
即 |AD'-AC'|=√5 (3)
AD'·BC'=AD'·(AC'-AB')=15
即AD'·AC'-AD'·AB'=15 (4)
由(2) 得
|AB'|²-2AB'·(AD'-AC')+|AD'-AC'|²=8
(1)(3)代入
1²-2AB'·(AD'-AC')+(√5)²=8
AB'·AC'-AB'·AD'=1 (5)
(4)-(5):
AD'·AC'-AB'·AC'=14
AC'·(AD'-AB')=AC'·BD'=14
所以 AC'·BD'=14
希望能帮到你!
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