标签：over math 解析 mat 因此 row lin 每日一题 spl
半径为\(1\)的圆上有三个动点\(A,B,C\),则\(\overrightarrow{AB}\cdot \overrightarrow{AC}\)的最小值为\((\qquad)\)
\(\mathrm{A}.-1\) \(\qquad\mathrm{B}.-\dfrac{3}{4}\) \(\qquad\mathrm{C}.-\dfrac{1}{2}\) \(\qquad\mathrm{D}.-\dfrac{1}{4}\)
解析:
不妨设\[
A(0,1),B(\cos\alpha,\sin\alpha),C(\cos\beta,\sin\beta),\alpha,\beta\in\left[0,2\pi\right).\]于是\[
\begin{split}
\overrightarrow{AB}\cdot\overrightarrow{AC}&=\left( \cos\alpha,\sin\alpha-1\right)\cdot \left(\cos\beta,\sin\beta-1\right)\ &=\cos\alpha\cos\beta+\sin\alpha\sin\beta-\left(\sin\alpha+\sin\beta\right)+1\ &=\cos\left(\alpha-\beta\right)-2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}+1\ &=2\cos\dfrac{\alpha-\beta}{2}\cdot \left(\cos\dfrac{\alpha-\beta}{2}-\sin\dfrac{\alpha+\beta}{2}\right)\ &\geqslant -2\cdot \left(\dfrac{1}{2}\sin\dfrac{\alpha+\beta}{2}\right)^2\ &\geqslant -\dfrac{1}{2}.
\end{split}
\]
因此当\(\cos\dfrac{\alpha-\beta}{2}=\dfrac{1}{2}\sin\dfrac{\alpha+\beta}{2}\)且\(\sin\dfrac{\alpha+\beta}{2}=1\)时,上述不等式取等.取\[(\alpha,\beta)=
\left(\dfrac{5\pi}{6},\dfrac{\pi}{6}\right).\]此时\(\overrightarrow{AB}\cdot\overrightarrow{AC}\)取得最小值\(-\dfrac{1}{2}\).
标签：over math 解析 mat 因此 row lin 每日一题 spl
原文地址：https://www.cnblogs.com/Math521/p/11871529.html
