Question

a向量=（2cosa,2sina)b向量=3(cosb,sinb)且a向量b向量夹角为60度，

a向量=（2cosa,2sina)b向量=3(cosb,sinb)且a向量b向量夹角为60度，求直线xcosa-ysina+1/2=0与园(x-cosb)^+(y+sinb)^=1的位置关系

Answer 1

解：
a*b=6（cosacosb+sinasinb）=|a||b|cos60°=6*1/2=3，
所以cosacosb+sinasinb=3/6=1/2，
(x-cosb)^+(y+sinb)^=1圆心的点为(cosb,-sinb),半径r=1，
则(cosb,-sinb)到xcosa-ysina+1/2=0的距离为
d=|cosbcosa+sinbsina+1/2|/√(cos²a+sin²a)
=1/1
=1
=r
所以直线xcosa-ysina+1/2=0与园(x-cosb)^+(y+sinb)^=1相切。

Answer 2

根据向量夹角公式ab=|a||b|cos60,60为a与b的夹角；
ab=（2cosa,2sina)乘以3(cosb,sinb)=6（cosacosb+sinasinb）;
|a||b|cosc=3,两边相等得：cosacosb+sinasinb=0.5（1式）；
根据点到直线的距离公式；圆是以（cosb，-sinb）为圆心的单位元，半径为1.
圆心到直线的距离为d=|cosacosb+sinasinb+0.5|/1=1(将1式代入得到)，
得圆心到直线的距离为1，正好为半径长，所以直线与圆的位置关系是相切。

Answer 3

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Answer 4

a·b=2cosA*3cosB+2sinA*3sinB=6(cosAcosB+sinAsinB)=6cos(A-B)
a·b=|a|*|b|cos60=2*3(1/2)=3
所以6cos(A-B)=3--->cos(A-B)=1/2
(x-cosB)^2+(y+sinB)^2=R? ? ?的圆心是P(cosB,-sinB)
点P到直线xcosA-ysinA+0.5=0的距离
d=|cosBcosA+sinBsinA+0.5|/√[(cosA)^2+(-sinA)^2]
=|cos(A-B)+1/2|
=|1/2+1/2|
=1
如果R=1,则直线与圆相切，如果R>1,则直线与圆相交，如果R<1,则直线与圆相离。