高一数学3已知A(-1,0)B(
已知A(-1,0)B(1,0)两点,C点在直线2x-3=0上,且AC*AB,CA*CB,BA*BC成等差数列,记θ为CA与CB的夹角,求tanθ
设坐标 C(3/2,y)
AC*AB = (3/2+1,y)*(1+1,0) = 5
BA*BC = (-1-1,0)*(3/2-1,0)=-1
CA*CB = (-1-3/2,y)*(1-3/2,y)=5/4+y^
AC*AB,CA*CB,BA*BC成等差数列,
--->2CA*CB = AC*AB + BA*BC
--->5/2+2y^=5-1=4--->y^=3/4--->y=±√3/2
--->|CB|=1,|CA|=√7
--->CA*...全部
已知A(-1,0)B(1,0)两点,C点在直线2x-3=0上,且AC*AB,CA*CB,BA*BC成等差数列,记θ为CA与CB的夹角,求tanθ
设坐标 C(3/2,y)
AC*AB = (3/2+1,y)*(1+1,0) = 5
BA*BC = (-1-1,0)*(3/2-1,0)=-1
CA*CB = (-1-3/2,y)*(1-3/2,y)=5/4+y^
AC*AB,CA*CB,BA*BC成等差数列,
--->2CA*CB = AC*AB + BA*BC
--->5/2+2y^=5-1=4--->y^=3/4--->y=±√3/2
--->|CB|=1,|CA|=√7
--->CA*CB = 5/4+3/4 = |CB||CA|cosθ
--->cosθ = 2/√7 --->sinθ=√3/√7
--->tanθ = √3/2。
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