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春芽
共回答了18个问题采纳率:83.3% 举报
(1)
a垂直b
=> a.b=0
(1,x).(2x+3,-x) =0
2x+3 -x^2 =0
x^2 -2x+3 =0
(x-3)(x+1)=0
x = 3 or -1
(2)
a//b
=> a = kb
(1,x) = k(2x+3,-x)
=>
2kx+3 =1 (1) and
x = -kx (2)
from (2) ,k=-1
from (1) ,x = 1
a-b = ( 1,1) -(5,-1)
=(-4,2)
|a-b| = 2√5
1年前
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