Dạng 2: Chứng minh hai đường thẳng vuông góc và các bài toán liên quan có đáp án

Chọn A.

a) Ta có $MC=MD=\frac{a\sqrt{3}}{2}$ nên tam giác MCD cân tại M , do đó $MN\perp CD$

Lại có $RP\parallel CD\Rightarrow MN\perp RQ$

b) Tương tự ta có $QP\perp AD$

Trong tam giác vuông PDQ  ta có $Q{P}^2=Q{D}^2-D{P}^2={\left(\frac{a\sqrt{3}}{2}\right)}^2-{\left(\frac{a}{2}\right)}^2=\frac{a^2}{2}$

Ta có $R{Q}^2+R{P}^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2=a^2=Q{P}^2$

Do đó tam giác RPQ  vuông tại R , hay $RP\perp RQ$

Vì vậy $\left\{\begin{array}{l}AB\parallel RQ\\ CD\parallel RP\\ RP\perp RQ\end{array}\right.\Rightarrow AB\perp CD$

a) Ta có $\left\{\begin{array}{l}\left(\alpha \right)\parallel \left(SAB\right)\\ \left(SAB\right)\cap \left(ABCD\right)=AB\\ \left(\alpha \right)\cap \left(ABCD\right)=MN\end{array}\right.$

Tương tự $\left\{\begin{array}{l}\left(\alpha \right)\parallel \left(SAB\right)\\ \left(SBC\right)\cap \left(SAB\right)=SB\\ \left(\alpha \right)\cap \left(SBC\right)=NP\end{array}\right.\Rightarrow NP\parallel SB$

$\left\{\begin{array}{l}\left(\alpha \right)\parallel \left(SAB\right)\\ \left(SAD\right)\cap \left(SAB\right)=SA\\ \left(\alpha \right)\cap \left(SAD\right)=MQ\end{array}\right.\Rightarrow MQ\parallel SA$

Dễ thấy $MN\parallel PQ\parallel AB\parallel CD$ nên MNPQ  là hình bình hành

Lại có $\left\{\begin{array}{l}MN\parallel AB\\ MQ\parallel SA\\ AB\perp SA\end{array}\right.\Rightarrow MN\perp MQ$

Vậy MNPQ  là hình thang vuông.

Chọn A

b) Ta có $MN=AB=a, MQ=\frac{SA}{2}=\frac{a}{2}, PQ=\frac{CD}{2}=\frac{a}{2}$

Vậy $S_{MNPQ}=\frac{1}{2}\left(MN+PQ\right).MQ=\frac{1}{2}\left(a+\frac{a}{2}\right)\frac{a}{2}=\frac{3a^2}{8}$

Đặt $\overrightarrow{AA\text{'}}=\overrightarrow{a},\overrightarrow{AB}=\overrightarrow{b},\overrightarrow{AD}=\overrightarrow{c}$

a) Ta có $\overrightarrow{AC\text{'}}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}, \overrightarrow{B\text{'}D\text{'}}=\overrightarrow{c}-\overrightarrow{b}$ 

nên $\overrightarrow{AC\text{'}}.\overrightarrow{B\text{'}D\text{'}}=\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)\left(\overrightarrow{c}-\overrightarrow{b}\right)$

$$\begin{gathered} =\overrightarrow{a}\left(\overrightarrow{c}-\overrightarrow{b}\right)+{\overrightarrow{c}}^2-{\overrightarrow{b}}^2=a^2-a^2=0 \\ \Rightarrow AC\text{'}\perp B\text{'}D\text{'} \end{gathered}$$

Chọn A

b) $\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=\left(\overrightarrow{AB}+\overrightarrow{BN}\right)-\left(\overrightarrow{AD}+\overrightarrow{DM}\right)$

$=\left(\overrightarrow{b}+\frac{x}{a}\overrightarrow{a}\right)-\left(\overrightarrow{c}+\frac{x}{a}\overrightarrow{b}\right)=\frac{x}{a}\overrightarrow{a}+\left(1-\frac{x}{a}\right)\overrightarrow{b}-\overrightarrow{c}$

Từ đó ta có $\overrightarrow{AC\text{'}}.\overrightarrow{MN}=\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)\text{[}\left(\overrightarrow{b}+\frac{x}{a}\overrightarrow{a}\right)-\left(\overrightarrow{c}+\frac{x}{a}\overrightarrow{b}\right)=\frac{x}{a}\overrightarrow{a}+\left(1-\frac{x}{a}\right)\overrightarrow{b}-\overrightarrow{c}\text{]}$

$=\frac{x}{a}{\overrightarrow{a}}^2+\left(1-\frac{x}{a}\right){\overrightarrow{b}}^2-{\overrightarrow{c}}^2=x.a+\left(1-\frac{x}{a}\right)a^2-a^2=0$

Vậy $AC\text{'}\perp MN$

Chọn A.

$B{C}^2=A{B}^2+A{C}^2-2AB.AC.\cos \left(AB,AC\right)=A{B}^2+A{C}^2-2.\overrightarrow{AB}.\overrightarrow{AC}$

Chọn B.

Ta có $\overrightarrow{AB}.\overrightarrow{EG}=\overrightarrow{AB}.\overrightarrow{AC}$, mặt khác $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}$

Chọn B.

Kẻ $NP\text{//A}C\left(P\in AB\right)$, nối MP

NP là đường trung bình $\Delta ABC\Rightarrow PN=\frac{1}{2}AC=\frac{a}{2}$

MP là đường trung bình $\Delta ABD\Rightarrow PM=\frac{1}{2}BD=\frac{3a}{2}$

Lại có $\left(AC,BD\right)=\left(PN,PM\right)=\overrightarrow{NPM}=90°$

=> tam giác MNP vuông tại P

Vậy $MN=\sqrt{P{N}^2+P{M}^2}=\frac{a\sqrt{10}}{2}$