作业帮一道高数向量积的计算题.设(a×b)⋅ c = 2,则[(a + b)×(b + c)] ⋅(c + a)=?我化简到[(a+b)b]⋅ (c+a)+ [(a+b)×c]⋅
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![一道高数向量积的计算题.设(a×b)⋅ c = 2,则[(a + b)×(b + c)] ⋅(c + a)=?我化简到[(a+b)b]⋅ (c+a)+ [(a+b)×c]⋅](/uploads/image/z/5169082-58-2.jpg?t=%E4%B8%80%E9%81%93%E9%AB%98%E6%95%B0%E5%90%91%E9%87%8F%E7%A7%AF%E7%9A%84%E8%AE%A1%E7%AE%97%E9%A2%98.%E8%AE%BE%28a%C3%97b%29%26%238901%3B+c+%3D+2%2C%E5%88%99%5B%28a+%2B+b%29%C3%97%28b+%2B+c%29%5D+%26%238901%3B%28c+%2B+a%29%3D%3F%E6%88%91%E5%8C%96%E7%AE%80%E5%88%B0%5B%28a%2Bb%29b%5D%26%238901%3B+%28c%2Ba%29%2B+%5B%28a%2Bb%29%C3%97c%5D%26%238901%3B)
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一道高数向量积的计算题.设(a×b)⋅ c = 2,则[(a + b)×(b + c)] ⋅(c + a)=?我化简到[(a+b)b]⋅ (c+a)+ [(a+b)×c]⋅
一道高数向量积的计算题.
设(a×b)⋅ c = 2,则[(a + b)×(b + c)] ⋅(c + a)=?
我化简到[(a+b)b]⋅ (c+a)+ [(a+b)×c]⋅
一道高数向量积的计算题.设(a×b)⋅ c = 2,则[(a + b)×(b + c)] ⋅(c + a)=?我化简到[(a+b)b]⋅ (c+a)+ [(a+b)×c]⋅
(a×b)⋅ c ＝（abc）＝（cab）＝（bca）,注意a×a＝0[即（cca）＝0等等]
[(a + b)×(b + c)] ⋅(c + a)
＝（abc）+（aba）+（acc）+（aca）＋（bbc）+（bba）+（bcc）+（bca）
＝（abc）+（abc）＝4