Bài toán vecto(2).

Question

1. Cho tứ diện $ABCD.$ Gọi $P,\,Q$ là các điểm thỏa $\overrightarrow{PA}=k\overrightarrow{PD},\,\overrightarrow{QB}=k\overrightarrow{QC}\,\,\,\left(k\neq1\right)$
       a) Biểu diễn $\overrightarrow{PQ}$ theo $\overrightarrow{AB},\,\overrightarrow{DC}.$
       b) Chứng minh (bằng vecto lớp 10) rằng: $PQ$ luôn song song với một mặt phẳng cố định.

2 (*). Cho hình hộp $ABCD.A'B'C'D'.$ Gọi $E,\,F,\,G,H$ là các điểm thỏa: $2\overrightarrow{ED}+\overrightarrow{EA}=\overrightarrow{0},\,2\overrightarrow{DF}+\overrightarrow{DD'}=\overrightarrow{0},\,2\overrightarrow{GF}+\overrightarrow{GC'}=\overrightarrow{0},\,3\overrightarrow{CH}+2\overrightarrow{CB}=\overrightarrow{0}.$
       a) Chứng minh: $E,\,G,\,H$ thẳng hàng.
       b) $M,\,N$ xác định bởi $3\overrightarrow{AM'}=\overrightarrow{A'G},\,\overrightarrow{A'N}=x\overrightarrow{A'E}.$ Xác định $x$ để $C',\,M,\,N$ thẳng hàng.

score 1 · Answer 1

2/a/
+$\begin{cases}2\overrightarrow{DF}=\overrightarrow{DD'} \\ 2\overrightarrow{GF}=\overrightarrow{GC'} \end{cases}\Rightarrow \begin{cases}\overrightarrow{FD}=\frac{1}{3}\overrightarrow{FD'} \\ \overrightarrow{FG}=\frac{1}{3}\overrightarrow{FC'} \end{cases}$
$\Rightarrow DG//D'C'$mặt khác $DC//D'C'$ nên G$\in DC$ hay E,G,H đồng phẳng (1)
+$DE//HC \Rightarrow \triangle DEG'\sim \triangle CHG'$ với $G'=EH\cap DC $
$\frac{DE}{CH}=\frac{DG'}{CG'}=\frac{EG'}{HG'}$ mặt khác $\frac{DE}{CH}=\frac{\frac{DA}{3}}{\frac{2}{3}BC}=\frac{1}{2}$
$\Rightarrow 2DG'=CG'$ (2)
$(1)\Rightarrow \triangle FDG\sim \triangle FD'C''$
$\Rightarrow \frac{DG}{D'C'}=\frac{FG}{FC'}=\frac{1}{3}\Rightarrow 3DG=D'C'$ hay $2DG=CG$ (3)
$(2)(3)=> G\equiv G'\Rightarrow E,G,H$ thẳng hàng (đpcm)

score 1 · Answer 2

Câu 1/a/
+$\overrightarrow{PA}=k\overrightarrow{PD}$
$\Leftrightarrow \overrightarrow{PD}+\overrightarrow{DA}=k\overrightarrow{PD}$
$\Leftrightarrow\overrightarrow{DA}=(k-1)\overrightarrow{PD}\Rightarrow \overrightarrow{AD}=(k-1)\overrightarrow{DP}$
Tương tự $\overrightarrow{BC}=(k-1)\overrightarrow{CQ}\Rightarrow \overrightarrow{BC}=(1-k)\overrightarrow{QC}$ (1)
+$\overrightarrow{AB}=\overrightarrow{AP}+\overrightarrow{PQ}+\overrightarrow{QB}$
Tương tự $\overrightarrow{DC}=\overrightarrow{DP}+\overrightarrow{PQ}+\overrightarrow{QC}$
$\Rightarrow \overrightarrow{AB}-\overrightarrow{DC}=\overrightarrow{AP}+\overrightarrow{PD}+\overrightarrow{QB}+\overrightarrow{CQ}=\overrightarrow{AD}+\overrightarrow{CB}$ (2)
+ $\overrightarrow{PQ}=\overrightarrow{PD}+\overrightarrow{DC}+\overrightarrow{CQ}$
$\overrightarrow{PQ}=\overrightarrow{PA}+\overrightarrow{AB}+\overrightarrow{BQ}$
$\Rightarrow 2 \overrightarrow{PQ}=\overrightarrow{PD}+\overrightarrow{DC}+\overrightarrow{CQ}+\overrightarrow{PA}+\overrightarrow{AB}+\overrightarrow{BQ}$ (3)
+ $\overrightarrow{PD}+\overrightarrow{CQ}+\overrightarrow{PA}+\overrightarrow{BQ}$
$=\overrightarrow{PD}+k\overrightarrow{PD}-(\overrightarrow{QC}+k\overrightarrow{QC})$
$=(k+1)\overrightarrow{PD}-(k+1)\overrightarrow{QC}$ (4)
$(1)(2)(4)\Rightarrow(4)= \frac{k+1}{1-k}\overrightarrow{PD}+\frac{k+1}{1-k}\overrightarrow{CB}=\frac{k+1}{1-k}(\overrightarrow{AB}-\overrightarrow{DC})$ (5)
Từ (3)(5) => $2\overrightarrow{PQ}=\frac{k+1}{1-k}(\overrightarrow{AB}-\overrightarrow{DC})+\overrightarrow{AB}+\overrightarrow{DC}$
$=\overrightarrow{PQ}=\frac{1}{1-k}(\overrightarrow{AB}+\overrightarrow{DC})$
b/
Cho I thuộc BD s/c : $\overrightarrow{IB}=k\overrightarrow{ID}$
Ta có $\frac{\overrightarrow{PA}}{\overrightarrow{IB}}=\frac{\overrightarrow{PD}}{\overrightarrow{ID}}\Rightarrow \overrightarrow{PI}}//{\overrightarrow{AB}$
$\Rightarrow {\overrightarrow{AB}//(PQI)}\Rightarrow \overrightarrow{AB}//\overrightarrow{PQ}$ => $\overrightarrow{PQ}$// mp chứa $\overrightarrow{AB}$