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利用向量积解三角形面积比例问题

下面我们利用向量外积解如下题目:

题目:已知$P$是平面$ABC$上任意一点,且满足$\lambda_1\overrightarrow{PA}+\lambda_2\overrightarrow{PB}+\lambda_3\overrightarrow{PC}=\mathbf{0}$,且$\lambda_1,\lambda_2,\lambda_3\neq 0$.记$\triangle ABP$,$\triangle BCP$,$\triangle ACP$的面积依次为$S_1,S_2,S_3$,求$S_1:S_2:S_3$的值.

:
$$
S_1=\frac{\left|\overrightarrow{PA}\times\overrightarrow{PB}\right|}{2},
$$
\begin{align*}
S_2&=\frac{\left|\overrightarrow{PB}\times\overrightarrow{PC}\right|}{2}
\\&=\frac{\left|\overrightarrow{PB}\times\left(\frac{-\lambda_{1}\overrightarrow{PA}-\lambda_{2}\overrightarrow{PB}}{\lambda_{3}}\right)\right|}{2}
\\&=\frac{\left|\frac{-\lambda_{1}}{\lambda_{3}}\overrightarrow{PB}\times
\overrightarrow{PA}-\frac{\lambda_2}{\lambda_3}\overrightarrow{PB}\times\overrightarrow{PB}\right|}{2}
\\&=\left|\frac{\lambda_1}{\lambda_3}\right|\frac{\left|\overrightarrow{PB}\times\overrightarrow{PA}\right|}{2}.
\end{align*}
同理可得
$$
S_3=\left|\frac{\lambda_2}{\lambda_3}\right|\frac{\left|\overrightarrow{PB}\times\overrightarrow{PA}\right|}{2}.
$$
因此$S_1:S_2:S_3=|\lambda_3|:|\lambda_1|:|\lambda_2|$.

推论:若$P$是三角形$ABC$的重心,则$S_1:S_2:S_3=1:1:1$.