2015年浙江理科数学第15题

青青子衿

15.已知\(\overrightarrow {{e_1}}\)，\(\overrightarrow {{e_2}} \)是空间单位向量，\(\overrightarrow {{e_1}}  \cdot \overrightarrow {{e_2}}  = \frac{1}{2}\)，若空间向量\(\overrightarrow b \)满足\(\overrightarrow b  \cdot \overrightarrow {{e_1}}  = 2\)，\(\overrightarrow b  \cdot \overrightarrow {{e_2}}  = \frac{5}{2}\)，且对于任意\(x\),\(y\)\(\in\)\(R\)，\(\left| {\overrightarrow b  - \left( {x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} } \right)} \right| \ge \left| {\overrightarrow b  - \left( {{x_0}\overrightarrow {{e_1}}  + {y_0}\overrightarrow {{e_2}} } \right)} \right| = 1\)\(\left( {{x_0},{y_0} \in R} \right)\)，则\(x_0\)=__________，\(y_0\)=__________，\(\left| {\overrightarrow b } \right|\)=__________．

hjfmhh

kuing

青青子衿

回复 2# hjfmhh

hjfmhh 发表于 2015-7-24 11:47

【解析】\(\because\overrightarrow {{e_1}}  \cdot \overrightarrow {{e_2}}  = \frac{1}{2}\)，
\(\therefore \left\langle {\overrightarrow {{e_1}} ,\overrightarrow {{e_2}} } \right\rangle  = \frac{\pi }{3}\)
设空间单位向量\(\overrightarrow {{e_1}}\)，\(\overrightarrow {{e_2}}\)分别为平面\(\alpha \)，\(\beta \)的单位法向量
如图，设\(\overrightarrow {OA}  = 2\overrightarrow {{e_1}} \)，\(\overrightarrow {OB}  = \frac{5}{2}\overrightarrow {{e_2}} \)，且点\(A\)，\(B\)分别在平面\(\alpha \)，\(\beta \)上，
设平面\(\alpha \)，\(\beta \)的交线为\(l\)，线段\({OA}\)，\({OB}\)所确定的平面为\(\gamma \)
\[\left\{ \begin{array}{l}
OA \bot \alpha \\
OB \bot \beta \\
\alpha  \cap \beta  = l
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
OA \bot l\\
OB \bot l\\
OA \cap OB = O
\end{array} \right. \Rightarrow l \bot \gamma \]
由于\(\overrightarrow {OA} \cdot \overrightarrow {{e_1}}  = 2{\left| {\overrightarrow {{e_1}} } \right|^2} = 2\)，\(\overrightarrow {OB}  \cdot \overrightarrow {{e_2}}  = \frac{5}{2}{\left| {\overrightarrow {{e_2}} } \right|^2} = \frac{5}{2}\)，
则\(\overrightarrow {OA}\)为\(\overrightarrow b \)在（平面\(\alpha \)的单位法向量）\(\overrightarrow {{e_1}} \)的方向上的分向量，
则\(\overrightarrow {OB}\)为\(\overrightarrow b \)在（平面\(\beta \)的单位法向量）\(\overrightarrow {{e_2}} \)的方向上的分向量，
若空间向量\(\overrightarrow b \)的起点为\(O\)，由于
空间向量\(\overrightarrow b \)的终点落在平面\(\alpha \)上，也落在平面\(\beta \)，则
空间向量\(\overrightarrow b \)的终点落在平面\(\alpha \)，\(\beta \)的交线\(l\)上，
记空间向量\(\overrightarrow b \)的终点为点\(P\)（点\(P\)为待定定点）
设交线\(l\)在平面\(\gamma \)上的垂足为\(Q\)，\(\angle AOQ = \theta \)，
（平面\(\alpha \)法向量）\(\overrightarrow {OA}\)与（平面\(\beta \)法向量）\(\overrightarrow {OB}\)的夹角等于\(\overrightarrow {{e_1}}\)与\(\overrightarrow {{e_2}} \)的夹角
则\(\overrightarrow b  = \overrightarrow {OP} \)（\(\overrightarrow b \)为待定常向量），\(\angle BOQ = \frac{\pi }{3} - \theta \)
\[\therefore \frac{{OA}}{{\cos \theta }} = \frac{{OB}}{{\cos \left( {\frac{\pi }{3} - \theta } \right)}} = OQ \Rightarrow \frac{2}{{\cos \theta }} = \frac{{\frac{5}{2}}}{{\cos \left( {\frac{\pi }{3} - \theta } \right)}}\]
\[\therefore \cos \theta  = \frac{{2\sqrt 7 }}{7},sin\theta  = \frac{{\sqrt {21} }}{7},\left| {\overrightarrow {OQ} } \right| = \frac{2}{{\cos \theta }} = \sqrt 7 \]
过点\(Q\)作\(CQ//OB\)交\({OA}\)于点\(C\)，
在\(RT\triangle AOQ\)中，\(AQ = OQ\sin \theta  = \sqrt 3 \)
在\(RT\triangle ACQ\)中，\(OC = OA - AC = 2 - \frac{{AQ}}{{\tan \angle ACQ}} = 2 - \frac{{AQ}}{{\tan \angle AOB}} = 1\)，
\[CQ = \frac{{AQ}}{{\sin \angle ACQ}} = \frac{{AQ}}{{\sin \angle AOB}} = 2\]
\[\overrightarrow {OQ}  = \overrightarrow {OC}  + \overrightarrow {CQ}  = \overrightarrow {{e_1}}  + 2\overrightarrow {{e_2}} \]
设\(\overrightarrow {OT}  = x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} \)，点\(T\)在平面\(\gamma \)上（点\(T\)为动点）
\[\left| {\overrightarrow b  - \left( {x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} } \right)} \right| = \left| {\overrightarrow {OP}  - \overrightarrow {OT} } \right| = \left| {\overrightarrow {TP} } \right| \ge {\left| {\overrightarrow {TP} } \right|_{\min }} = \left| {\overrightarrow {QP} } \right| = 1\]，
\[\left| {\overrightarrow b  - \left( {x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} } \right)} \right| \ge \left| {\overrightarrow b  - \left( {{x_0}\overrightarrow {{e_1}}  + {y_0}\overrightarrow {{e_2}} } \right)} \right| = \left| {\overrightarrow {OP}  - \overrightarrow {OQ} } \right| = \left| {\overrightarrow {QP} } \right| = 1\]，
\[\therefore {x_0} = 1,{y_0} = 2,\left| {\overrightarrow b } \right| = \left| {\overrightarrow {OP} } \right| = \sqrt {{{\left| {\overrightarrow {OQ} } \right|}^2} + {{\left| {\overrightarrow {QP} } \right|}^2}}  = 2\sqrt 2 \]

回复 3# kuing

kuing 发表于 2015-7-24 12:55

【答案】\(1\)，\(2\)，\(2\sqrt 2 \)
【解析】问题等价于：\(\left| {\overrightarrow b  - \left( {x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} } \right)} \right|\)当且仅当\(x = {x_0}\)，\(y = {y_0}\)时，取得最小值\(1\)，
将\(\left| {\overrightarrow b  - \left( {x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} } \right)} \right|\)平方得：
\[{\left| {\overrightarrow b } \right|^2} - 2\overrightarrow b  \cdot \left( {x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} } \right) + {\left( {x\overrightarrow {{e_1}}  + y\overrightarrow {{e_2}} } \right)^2} = {\left| {\overrightarrow b } \right|^2} - \left( {4x + 5y} \right) + {x^2} + {y^2} + xy\]
\({\left| {\overrightarrow b } \right|^2} - \left( {4x + 5y} \right) + {x^2} + {y^2} + xy\)当且仅当\(x = {x_0}\)，\(y = {y_0}\)时，取得最小值\(1\)，
\[\begin{array}{l}
{\left| {\overrightarrow b } \right|^2} - \left( {4x + 5y} \right) + {x^2} + {y^2} + xy = {x^2} + \left( {y - 4} \right)x + {y^2} - 5y + {\left| {\overrightarrow b } \right|^2}\\
= {x^2} + \left( {y - 4} \right)x + \frac{{{{\left( {y - 4} \right)}^2}}}{4} - \frac{{{{\left( {y - 4} \right)}^2}}}{4} + {y^2} - 5y + {\left| {\overrightarrow b } \right|^2}\\
= {\left( {x + \frac{{y - 4}}{2}} \right)^2} + \frac{3}{4}{y^2} - 3y - 4 + {\left| {\overrightarrow b } \right|^2} = {\left( {x + \frac{{y - 4}}{2}} \right)^2} + \frac{3}{4}{y^2} - 3y + 3 - 7 + {\left| {\overrightarrow b } \right|^2}\\
= {\left( {x + \frac{{y - 4}}{2}} \right)^2} + \frac{3}{4}{\left( {y - 2} \right)^2} - 7 + {\left| {\overrightarrow b } \right|^2}
\end{array}\]
\[\therefore \left\{ \begin{array}{l}
{x_0} + \frac{{{y_0} - 4}}{2} = 0\\
{y_0} - 2 = 0\\
- 7 + {\left| {\overrightarrow b } \right|^2} = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_0} = 1\\
{y_0} = 2\\
\left| {\overrightarrow b } \right| = 2\sqrt 2
\end{array} \right.\]
考点：1.平面向量的模长；2.函数的最值

kuing

回复 4# 青青子衿

有必要重复写一次么……还写这么长，明明三行就可以完事……

其妙

回复 5# kuing
人家追求完美，做到细节，值得表彰！
哪像扎克，此处显然可得。。。。