问题补充:
边长为根号2的等边三角形ABC中,设向量AB=C,BC=a,CA=b,则向量a•b+c•a 等于
答案:
a·b=|BC|*|CA|*cos(π-C) =-根号2^2cos(π/3) =-根号2^2/2c·a=|AB|*|BC|*cos(π-B) =-根号2^2cos(π/3) =-根号2^2/2
故向量ab+ca=-根号2^2
时间:2024-07-13 11:56:53
边长为根号2的等边三角形ABC中,设向量AB=C,BC=a,CA=b,则向量a•b+c•a 等于
a·b=|BC|*|CA|*cos(π-C) =-根号2^2cos(π/3) =-根号2^2/2c·a=|AB|*|BC|*cos(π-B) =-根号2^2cos(π/3) =-根号2^2/2
故向量ab+ca=-根号2^2
已知实数a b c 满足ab+bc+ca=1 求证a根号bc+b根号ac+c根号ab
2018-11-05