上海市2020届高考数学一轮复习 专题突破训练 平面向量 文

?sin??cos???272 ① 25由(sin??cos?)?1?2sin?cos??98527?0, ,得2sin?cos??6256253???(?,?)

2?sin??cos???1?2sin?cos???242 ② 25由①、②得sin???312 50urrr2、解:(1)设n?(x,y).由m?n??1,得x?y??1 ①

rur3?22又向量n与向量m的夹角为,得x?y?1 ②

4rr?x??1?x?0由①、②解得?或?,?n?(?1,0)或n?(0,?1)

y?0y??1?? rrr(2)向量n与q?(1,0)共线知n?(?1,0);

由2B?A?C知B?

?3,2?2? A?C?,0?A?33rur?C?n?p???1?2cos2,cosA???cosC,cosA?,

2??rur21?cos2A1?cos2C ?n?p?cos2C?cos2A??221?1???4?????1??cos2A?cos??2A???1?cos?2A??

2?23??3???Q0?A?2???5???1?,?2A??,??1?cos?2A???, 33333?2?rur2?15?1??5?得?1?cos?2A???,即n?p??,?, 23?4??24?rur?25??n?p??,?? 22??