Question

若向量a,b满足|a|=|b|=1，a垂直b，且（2a+3b）垂直(ka-4b

若向量a,b满足|a|=|b|=1，a垂直b，且（2a+3b）垂直(ka-4b)求实数k的值

Answer 1

因为向量(ka-4b)和(2a+3b)垂直,所以(ka-4b)*(2a+3b)= 0,
(ka-4b)*(2a+3b)=2ka*a + 3ka*b-8a*b-12b*b, 注意到条件|a|=|b|=1,
则a*a = 1, b*b=1; 而a垂直于b,所以a*b=0. 所以,2k-12=0,k=6.

Answer 2

2a+3b与b夹角为tan2/3,若为直角,则ka-4b应与-b夹角为tan3/2,即k/4=3/2,得k=6