Question

在等边△ABC中，=a,=b,=c,|a|=2,

(1)求证：a⊥(b-c);

(2)解关于x的不等式|xa+b+c|＞1.

Answer 1

(1)证明：∵△ABC为等边三角形，且=a,=b,=c,|a|=2,

∴|a|=|b|=|c|=2,且a、b、c两两夹角均为120°.                                  

故a·(b-c)=a·b-a·c=|a||b|cos120°-|a||c|cos120°=0.                            

∴a⊥(b-c).                                                                 

(2)解：原不等式可化为|xa+b+c|²＞1,                                          

即a²x²+b²+c²+2(a·b+a·c)x+2b·c＞1,

即4x²-8x+3＞0.                                                            

解得x＜或x＞.

∴原不等式的解集为{x|x＜或x＞}.