题目内容

4.正四面体ABCD边长为a,点E、F分别是BC、AD的中点,则$\begin{array}{l}→\\{AE}\end{array}•\begin{array}{l}→\\{AF}\end{array}$的值为(  )
A.a2B.$\frac{1}{2}{a^2}$C.$\frac{1}{4}{a^2}$D.$\frac{{\sqrt{3}}}{4}{a^2}$

分析 如图所示,$\overrightarrow{AF}=\frac{1}{2}\overrightarrow{AD}$,$\overrightarrow{AE}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})$,代入$\begin{array}{l}→\\{AE}\end{array}•\begin{array}{l}→\\{AF}\end{array}$,利用数量积运算性质即可得出.

解答 解:如图所示,
$\overrightarrow{AF}=\frac{1}{2}\overrightarrow{AD}$,$\overrightarrow{AE}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})$,
∴$\begin{array}{l}→\\{AE}\end{array}•\begin{array}{l}→\\{AF}\end{array}$=$\frac{1}{4}$$(\overrightarrow{AD}•\overrightarrow{AB}+\overrightarrow{AD}•\overrightarrow{AC})$
=$\frac{1}{4}$(a2cos60°+a2cos60°)
=$\frac{1}{4}{a}^{2}$.
故选:C.

点评 本题考查了数量积运算性质、向量平行四边形法则,考查了推理能力与计算能力,属于中档题.

练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网