13 câu Trắc nghiệm Ôn tập Toán 10 Chương 2 Hình học có đáp án (Thông hiểu)

Đáp án D

Ta có: $\overrightarrow{BA}=\left(3;-1\right);\overrightarrow{BC}=\left(-4;-2\right)$

Suy ra

$$\begin{gathered} \cos \left(\overrightarrow{BA},\overrightarrow{BC}\right)=\frac{\overrightarrow{BA}.\overrightarrow{BC}}{\left|\overrightarrow{BA}\right|.\left|\overrightarrow{BC}\right|} \\ =\frac{3.(-4)+(-1).(-2)}{\sqrt{9+1}.\sqrt{16+4}}=-\frac{\sqrt{2}}{2} \\ \Rightarrow \widehat{B}=\left(\overrightarrow{BA};\overrightarrow{BC}\right)={135}^0 \end{gathered}$$

Đáp án D

Vì G và G’ lần lượt là trọng tâm của tam giác ABC và A’B’C’ nên ta có:

$$\begin{gathered} \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0} \\ \overrightarrow{G\text{'}A\text{'}}+\overrightarrow{G\text{'}B\text{'}}+\overrightarrow{G\text{'}C\text{'}}=\overrightarrow{0} \end{gathered}$$

Với điểm M bất kì khác điểm G ta chứng minh $3\overrightarrow{MG}=\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}$

Ta có:

$\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}=3\overrightarrow{MG}$

Tương tự ta có: $3\overrightarrow{MG\text{'}}=\overrightarrow{MA\text{'}}+\overrightarrow{MB\text{'}}+\overrightarrow{MC\text{'}}$

Từ đó suy ra

$$\begin{gathered} 3\overrightarrow{GG\text{'}}=3\left(\overrightarrow{MG\text{'}}-\overrightarrow{MG}\right)=3\overrightarrow{MG\text{'}}-3\overrightarrow{MG} \\ =\overrightarrow{MA\text{'}}+\overrightarrow{MB\text{'}}+\overrightarrow{MC\text{'}}-\overrightarrow{MA}-\overrightarrow{MB}-\overrightarrow{MC} \\ =\left(\overrightarrow{MA\text{'}}-\overrightarrow{MA}\right)+\left(\overrightarrow{MB\text{'}}-\overrightarrow{MB}\right)+\left(\overrightarrow{MC\text{'}}-\overrightarrow{MC}\right) \\ =\overrightarrow{AA\text{'}}+\overrightarrow{BB\text{'}}+\overrightarrow{CC\text{'}} \end{gathered}$$

Nên A đúng

Đáp án B:

$$\begin{gathered} 3\overrightarrow{GG\text{'}}=\overrightarrow{GG\text{'}}+\overrightarrow{GG\text{'}}+\overrightarrow{GG\text{'}} \\ =\overrightarrow{GA}+\overrightarrow{AC\text{'}}+\overrightarrow{C\text{'}G\text{'}}+\overrightarrow{GB}+\overrightarrow{BA\text{'}}+\overrightarrow{A\text{'}G\text{'}}+\overrightarrow{GC}+\overrightarrow{CB\text{'}}+\overrightarrow{B\text{'}G\text{'}} \\ =\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)+\left(\overrightarrow{AC\text{'}}+\overrightarrow{BA\text{'}}+\overrightarrow{CB\text{'}}\right)+\left(\overrightarrow{C\text{'}G\text{'}}+\overrightarrow{A\text{'}G\text{'}}+\overrightarrow{B\text{'}G\text{'}}\right) \\ =\overrightarrow{0}+\left(\overrightarrow{AC\text{'}}+\overrightarrow{BA\text{'}}+\overrightarrow{CB\text{'}}\right)+\overrightarrow{0} \\ =\overrightarrow{AC\text{'}}+\overrightarrow{BA\text{'}}+\overrightarrow{CB\text{'}} \end{gathered}$$

Nên B đúng

Đáp án C:

$$\begin{gathered} 3\overrightarrow{GG\text{'}}=\overrightarrow{GG\text{'}}+\overrightarrow{GG\text{'}}+\overrightarrow{GG\text{'}} \\ =\overrightarrow{GA}+\overrightarrow{AB\text{'}}+\overrightarrow{B\text{'}G\text{'}}+\overrightarrow{GB}+\overrightarrow{BC\text{'}}+\overrightarrow{C\text{'}G\text{'}}+\overrightarrow{GC}+\overrightarrow{CA\text{'}}+\overrightarrow{A\text{'}G\text{'}} \\ =\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)+\left(\overrightarrow{AB\text{'}}+\overrightarrow{BC\text{'}}+\overrightarrow{CA\text{'}}\right)+\left(\overrightarrow{B\text{'}G\text{'}}+\overrightarrow{C\text{'}G\text{'}}+\overrightarrow{A\text{'}G\text{'}}\right) \\ =\overrightarrow{0}+\left(\overrightarrow{AB\text{'}}+\overrightarrow{BC\text{'}}+\overrightarrow{CA\text{'}}\right)+\overrightarrow{0} \\ =\overrightarrow{AB\text{'}}+\overrightarrow{BC\text{'}}+\overrightarrow{CA\text{'}} \end{gathered}$$

Nên C đúng

D sai do A đúng

Đáp án B

Ta có: $\overrightarrow{AB}=\left(3;3\right),\overrightarrow{AC}=\left(7;-1\right),\overrightarrow{BC}=\left(4;-4\right)$

Do $\overrightarrow{AB}.\overrightarrow{BC}=3.4+3.(-4)=0$ nên $AB\perp BC$ hay tam giác ABC vuông tại B

Đáp án B

Ta có:

$$\begin{gathered} \overrightarrow{CI}=\overrightarrow{CA}-\overrightarrow{IA}=\overrightarrow{CA}-3\overrightarrow{IB} \\ =\overrightarrow{CA}-3\left(\overrightarrow{CB}-\overrightarrow{CI}\right) \\ \iff \overrightarrow{CI}=\overrightarrow{CA}-3\overrightarrow{CB}+3\overrightarrow{CI} \\ \iff 2\overrightarrow{CI}=3\overrightarrow{CB}-\overrightarrow{CA} \\ \iff \overrightarrow{CI}=\frac{1}{2}\left(3\overrightarrow{CB}-\overrightarrow{CA}\right) \end{gathered}$$

Đáp án D

Lấy điểm D sao cho $\overrightarrow{AB}=\overrightarrow{BD}\Rightarrow \left(\overrightarrow{AB},\overrightarrow{BC}\right)=\left(\overrightarrow{BD};\overrightarrow{BC}\right)={120}^0$

$$\begin{gathered} \Rightarrow \overrightarrow{AB}.\overrightarrow{BC}=\left|\overrightarrow{AB}\right|.\left|\overrightarrow{BC}\right|.\cos \left(\overrightarrow{AB},\overrightarrow{BC}\right) \\ =\mathrm{1.1.}\cos {120}^0=-\frac{1}{2} \end{gathered}$$

(vì tam giác ABC đều cạnh a nên AB = BC = 1$\Rightarrow \left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=1$)

Đáp án B

Ta có:

$$\begin{gathered} \left(\overrightarrow{IA}+\overrightarrow{IB}\right).\overrightarrow{ID}=\left(\overrightarrow{IA}+\overrightarrow{IA}+\overrightarrow{AB}\right).\overrightarrow{ID} \\ =2\overrightarrow{IA}.\overrightarrow{ID}=-\frac{9a^2}{2} \end{gathered}$$

Đáp án C

Phương án A: do $\overrightarrow{AB}=\left(2;2\right)$ nên A loại

Phương án B: $\overrightarrow{AB}=\left(2;2\right);\overrightarrow{BC}=\left(0;-4\right);\overrightarrow{AB}.\overrightarrow{BC}=-8$ suy ra $\overrightarrow{AB}$ không vuông góc với $\overrightarrow{BC}$ nên loại B

Phương án C: Ta có: $\overrightarrow{AB}=\left(2;2\right);\overrightarrow{AC}=\left(2;-2\right);\overrightarrow{BC}=\left(0;-4\right)$  suy ra $AB = AC =\sqrt{8}, \overrightarrow{AB}.\overrightarrow{AC}=0$

Nên tam giác ABC vuông cân tại A. Do đó chọn C

Đáp án C

$$\begin{gathered} \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1;\left|\overrightarrow{a}+\overrightarrow{b}\right|=2 \\ \iff {\left(\overrightarrow{a}+\overrightarrow{b}\right)}^2=4\iff \overrightarrow{a}.\overrightarrow{b}=1 \\ \left(3\overrightarrow{a}-4\overrightarrow{b}\right)\left(2\overrightarrow{a}+5\overrightarrow{b}\right)=6{\overrightarrow{a}}^2-20{\overrightarrow{b}}^2+7\overrightarrow{a}.\overrightarrow{b}=-7 \end{gathered}$$

Đáp án A

Ta có:

$$\begin{gathered} \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{{\left(\overrightarrow{a}+\overrightarrow{b}\right)}^2}=\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2+2\overrightarrow{a}.\overrightarrow{b}} \\ =\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\overrightarrow{a},\overrightarrow{b}\right)}=\sqrt{21} \end{gathered}$$

Đáp án C

Ta có: $\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{0}$

$$\begin{gathered} \iff \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=-2\overrightarrow{c}\iff {\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)}^2=4{\overrightarrow{c}}^2 \\ \iff a^2+b^2+c^2+2A=4c^2 \end{gathered}$$

Do đó: $A=\frac{1}{2}\left(3c^2-a^2-b^2\right)$

Đáp án C

Từ giả thiết suy ra $AC=a\sqrt{2}$

Ta có: $P=\overrightarrow{AC}.\left(\overrightarrow{CD}+\overrightarrow{CA}\right)$

$$\begin{gathered} =\overrightarrow{AC}.\overrightarrow{CD}+\overrightarrow{AC}.\overrightarrow{CA}=-\overrightarrow{CA}.\overrightarrow{CD}-{\overrightarrow{AC}}^2 \\ =-CA.CD.\cos \left(\overrightarrow{CA},\overrightarrow{CD}\right)-A{C}^2 \\ =-a\sqrt{2}.a.\cos {45}^0-{\left(a\sqrt{2}\right)}^2=-3a^2 \end{gathered}$$

Đáp án D

Phương án A: $\overrightarrow{AB}=\left(-3;-2\right),\overrightarrow{AC}=\left(3;-\frac{9}{2}\right)$ nên A đúng

Phương án B: $\overrightarrow{AB}.\overrightarrow{AC}=0$ nên B đúng

Phương án C: $\left|\overrightarrow{AB}\right|=\sqrt{13}$ nên C đúng

Phương án D: Ta có $\overrightarrow{BC}=\left(6;-\frac{5}{2}\right)$ suy ra $BC=\sqrt{6^2+{\left(\frac{5}{2}\right)}^2}=\frac{13}{2}$ nên D sai

Đáp án D

Ta có:

$\overrightarrow{AB}=\left(8;4\right),\overrightarrow{AD}=\left(5;-5\right),\overrightarrow{CB}=\left(-2;4\right),\overrightarrow{CD}=\left(-5;-5\right)$

Suy ra $\left\{\begin{array}{c}\cos \left(\overrightarrow{AB},\overrightarrow{AD}\right)=\frac{8.5+4.(-5)}{\sqrt{8^2+4^2}\sqrt{5^2+5^2}}=\frac{1}{\sqrt{10}}\\ \cos \left(\overrightarrow{CB},\overrightarrow{CD}\right)=\frac{(-2).(-5)+4.(-5)}{\sqrt{2^2+4^2}\sqrt{5^2+5^2}}=-\frac{1}{\sqrt{10}}\end{array}\right.$

Suy ra

$$\begin{gathered} \cos \left(\overrightarrow{AB},\overrightarrow{AD}\right)+\cos \left(\overrightarrow{CB},\overrightarrow{CD}\right)=0 \\ \Rightarrow \widehat{BAD}+\widehat{BCD}={180}^0 \end{gathered}$$