Bài tập Tích vô hướng của hai vecto có đáp án

Tích vô hướng của hai vecto $\overrightarrow{u},\overrightarrow{v}\ne \overrightarrow{0}$ được tính bởi công thức sau:

$\overrightarrow{u}.\overrightarrow{v}=\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|.c\text{os}\left(\overrightarrow{u},\overrightarrow{v}\right).$

Vì $\left|\overrightarrow{u}\right|>0,\left|\overrightarrow{v}\right|>0$ nên dấu của $\overrightarrow{u}.\overrightarrow{v}$ phụ thuộc vào dấu của $c\text{os}\left(\overrightarrow{u},\overrightarrow{v}\right)$.

Nếu tích vô hướng của hai vecto $\overrightarrow{u},\overrightarrow{v}$ là một số dương thì $c\text{os}\left(\overrightarrow{u},\overrightarrow{v}\right)>0.$ Do đó góc giữa hai vecto $\overrightarrow{u},\overrightarrow{v}$ là góc nhọn hoặc bằng 0⁰.

Nếu tích vô hướng của hai vecto $\overrightarrow{u},\overrightarrow{v}$ là một số âm thì $c\text{os}\left(\overrightarrow{u},\overrightarrow{v}\right)<0.$ Do đó góc giữa hai vecto $\overrightarrow{u},\overrightarrow{v}$ là góc tù hoặc bằng 180⁰.

a) Ta có: $\overrightarrow{u}=\overrightarrow{0}\Rightarrow \left\{\begin{array}{l}x=0\\ y=0\end{array}\right.$

Mà $\overrightarrow{0}$ vuông góc với mọi vecto nên ta có: $\overrightarrow{u}.\overrightarrow{v}=0$

Ta lại có: $k\left(x^2+y^2\right)=k\left(0^2+0^2\right)=0$

$\Rightarrow \overrightarrow{u}.\overrightarrow{v}=k\left(x^2+y^2\right)$

Vậy với $\overrightarrow{u}=\overrightarrow{0}$ công thức đã cho đúng.

b) Vì k ≥ 0 nên hai vecto $\overrightarrow{u},\overrightarrow{v}$ cùng hướng

$\Rightarrow \left(\overrightarrow{u},\overrightarrow{v}\right)=0^0$

Ta có: 

$\overrightarrow{u}.\overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|cos\left(\overrightarrow{u},\overrightarrow{v}\right)=\sqrt{x^2+y^2}.\sqrt{{\left(kx\right)}^2+{\left(ky\right)}^2}\text{.cos}\left(\overrightarrow{u},\overrightarrow{v}\right)=\left|k\right|\left(x^2+y^2\right).c\text{os}{0}^0=k\left(x^2+y^2\right).$

Vậy với $\overrightarrow{u}\ne \overrightarrow{0}$ và $k\ge 0$ công thức đã cho đúng.

c) Vì k < 0 nên hai vecto $\overrightarrow{u},\overrightarrow{v}$ ngược hướng

$\Rightarrow \left(\overrightarrow{u},\overrightarrow{v}\right)={180}^0$

Ta có:

$\overrightarrow{u}.\overrightarrow{v}=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|cos\left(\overrightarrow{u},\overrightarrow{v}\right)=\sqrt{x^2+y^2}.\sqrt{{\left(kx\right)}^2+{\left(ky\right)}^2}\text{.cos}\left(\overrightarrow{u},\overrightarrow{v}\right)=\left|k\right|\left(x^2+y^2\right).c\text{os18}{0}^0$

Vậy với $\overrightarrow{u}\ne \overrightarrow{0}$ và k < 0 công thức đã cho đúng.

a) Vì $\overrightarrow{OA}=\overrightarrow{u}$ mà $\overrightarrow{u}=\left(x;y\right)$ nên  $\overrightarrow{OA}=\left(x;y\right)$ suy ra A(x; y).

Vì $\overrightarrow{OB}=\overrightarrow{v}$ mà $\overrightarrow{v}=\left(x\text{'};y\text{'}\right)$ nên  $\overrightarrow{OB}=\left(x^{\text{'}};y^{\text{'}}\right)$ suy ra B(x'; y').

b) +) Ta có: A(x; y) và B(x'; y')$\Rightarrow \overrightarrow{AB}=\left(x\text{'}-x;y\text{'}-y\right)$

 $\Rightarrow AB=\sqrt{{\left(x\text{'}-x\right)}^2+{\left(y\text{'}-y\right)}^2}$

 $\Rightarrow A{B}^2={\left(x\text{'}-x\right)}^2+{\left(y\text{'}-y\right)}^2.$

+) Ta có: $\overrightarrow{OA}=\left(x;y\right)\Rightarrow OA=\sqrt{x^2+y^2}\Rightarrow O{A}^2=x^2+y^2.$

Vậy $A{B}^2={\left(x\text{'}-x\right)}^2+{\left(y\text{'}-y\right)}^2;$ $O{A}^2=x^2+y^2$ và $O{B}^2=x{\text{'}}^2+y{\text{'}}^2.$

c) Ta có:  $\overrightarrow{OA}.=OA.OB.cos\left(\overrightarrow{OA},\overrightarrow{OB}\right)=OA.OB.cos\widehat{AOB}$

Xét tam giác OAB, theo định lí côsin ta có: $cos\widehat{AOB}=\frac{O{A}^2+O{B}^2-A{B}^2}{2.OA.OB}$

 $\Rightarrow cos\widehat{AOB}=\frac{x^2+y^2+{x^{\text{'}}}^2+{y^{\text{'}}}^2-\left[{\left(x^{\text{'}}-x\right)}^2+{\left(y^{\text{'}}-y\right)}^2\right]}{2.\sqrt{x^2+y^2}.\sqrt{{x^{\text{'}}}^2+{y^{\text{'}}}^2}}$

$$\begin{gathered} \Rightarrow cos\widehat{AOB}=\frac{x^2+y^2+{x^{\text{'}}}^2+{y^{\text{'}}}^2-\left({x^{\text{'}}}^2-2x^{\text{'}}x+x^2+{y^{\text{'}}}^2-2y^{\text{'}}y+y^2\right)}{2.\sqrt{x^2+y^2}.\sqrt{{x^{\text{'}}}^2+{y^{\text{'}}}^2}} \\ \Rightarrow cos\widehat{AOB}=\frac{x^2+y^2+{x^{\text{'}}}^2+{y^{\text{'}}}^2-{x^{\text{'}}}^2+2x^{\text{'}}x-x^2-{y^{\text{'}}}^2+2y^{\text{'}}y-y^2}{2.\sqrt{x^2+y^2}.\sqrt{{x^{\text{'}}}^2+{y^{\text{'}}}^2}} \\ \Rightarrow cos\widehat{AOB}=\frac{2x^{\text{'}}x+2y^{\text{'}}y}{2.\sqrt{x^2+y^2}.\sqrt{{x^{\text{'}}}^2+{y^{\text{'}}}^2}} \\ \Rightarrow cos\widehat{AOB}=\frac{2.\left(x^{\text{'}}x+y^{\text{'}}y\right)}{2.\sqrt{x^2+y^2}.\sqrt{{x^{\text{'}}}^2+{y^{\text{'}}}^2}} \\ \Rightarrow cos\widehat{AOB}=\frac{x^{\text{'}}x+y^{\text{'}}y}{\sqrt{x^2+y^2}.\sqrt{{x^{\text{'}}}^2+{y^{\text{'}}}^2}} \end{gathered}$$

Do đó $\overrightarrow{OA}.\overrightarrow{OB}=OA.OB.cos\widehat{AOB}$

Vậy $\overrightarrow{OA}.\overrightarrow{OB}=x^{\text{'}}x+y^{\text{'}}y$.

a) Vì H là trực tâm của tam giác ABC nên $AH\perp BC\Rightarrow \overrightarrow{AH}.\overrightarrow{BC}=0$

và $BH\perp AC\Rightarrow \overrightarrow{BH}.\overrightarrow{AC}=0$

$BH\perp CA\Rightarrow \overrightarrow{BH}\perp \overrightarrow{CA}\Rightarrow \overrightarrow{BH}.\overrightarrow{CA}=0$

b) Gọi tọa độ điểm H(x;y), ta có:

$\overrightarrow{AH}\left(x+1;y-2\right),\overrightarrow{BH}\left(x-8;y+1\right),\overrightarrow{BC}\left(0;9\right),\overrightarrow{AC}\left(9;6\right)$

$Suy ra \stackrel{\rightharpoonup }{AH}.\overrightarrow{BC}=\left(x+1\right).0+\left(y-2\right).9=0\iff 9(y-2)$

$Và \overrightarrow{BH}.\overrightarrow{AC}=\left(x-8\right).9+\left(y+1\right).6=9x+6y-66$

⇔ 9x - 54=0

$\iff$9x= 54

⇔ x = 6

⇒ H(6; 2)

Vậy H(6;2).

c)Với A(‒1;2), B(8;‒1), C(8;8) ta có:

$\overrightarrow{AB}=\left(9;-3\right)\Rightarrow AB=\sqrt{9^2+{\left(-3\right)}^2}=3\sqrt{10}.$

$\overrightarrow{AC}\left(9;6\right)\Rightarrow AC=\sqrt{9^2+6^2}=3\sqrt{13}.$

$\overrightarrow{BC}\left(0;9\right)\Rightarrow BC=\sqrt{0^2+9^2}=9.$

$cos\widehat{BAC}=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{AB.AC}=\frac{63}{3\sqrt{10}.3\sqrt{13}}=\frac{63}{9.\sqrt{130}}=\frac{7}{\sqrt{130}}$

a) Ta có: $\overrightarrow{a}.\overrightarrow{b}=\left(-3\right).2+1.6=0\Rightarrow \left(\overrightarrow{a},\overrightarrow{b}\right)={90}^0.$

b)  Ta có: $\overrightarrow{a}.\overrightarrow{b}=3.2+1.4=10$

$\left|\overrightarrow{a}\right|=\sqrt{3^2+1^2}=\sqrt{10},\left|\overrightarrow{b}\right|=\sqrt{2^2+4^2}=2\sqrt{5}$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)\Rightarrow c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{10}{\sqrt{10}.2\sqrt{5}}=\frac{1}{\sqrt{2}}\Rightarrow \left(\overrightarrow{a},\overrightarrow{b}\right)={45}^0.$

c) Ta có: $\overrightarrow{a}.\overrightarrow{b}=\left(-\sqrt{2}\right).2+1.\left(-\sqrt{2}\right)=-3\sqrt{2}$

$\left|\overrightarrow{a}\right|=\sqrt{{\left(-\sqrt{2}\right)}^2+1^2}=\sqrt{3},\left|\overrightarrow{b}\right|=\sqrt{2^2+{\left(-\sqrt{2}\right)}^2}=\sqrt{6}$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)\Rightarrow c\text{os}\left(\overrightarrow{a},\overrightarrow{b}\right)=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{-3\sqrt{2}}{\sqrt{3}.\sqrt{6}}=-1\Rightarrow \left(\overrightarrow{a},\overrightarrow{b}\right)={180}^0.$

a) Ta có: $\overrightarrow{AM}\left(t-1;-2\right),\overrightarrow{BM}\left(t+4;-3\right)$

$\Rightarrow \overrightarrow{AM}.\overrightarrow{BM}=\left(t-1\right)\left(t+4\right)+\left(-2\right).\left(-3\right)=t^2+3t+2.$.

b) Để $\widehat{AMB}={90}^0$ thì $\overrightarrow{AM}.\overrightarrow{BM}=0$

$\iff t^2+3t+2=0\iff \left[\begin{array}{l}t=-1\\ t=-2\end{array}\right.$

Vậy với t = -1 hoặc t = - 2 thì $\widehat{AMB}={90}^0.$

a) Ta có: $\overrightarrow{AB}\left(6;3\right)\Rightarrow AB=\sqrt{6^2+3^2}=3\sqrt{5};$

$\overrightarrow{AC}\left(6;-3\right)\Rightarrow AC=\sqrt{6^2+{\left(-3\right)}^2}=3\sqrt{5};$

$\overrightarrow{BC}\left(0;-6\right)\Rightarrow BC=\sqrt{0^2+{\left(-6\right)}^2}=6;$

Theo định lí cosin, ta có:

$\text{cos}A=\frac{A{B}^2+A{C}^2-B{C}^2}{2.AB.AC}=\frac{3}{5}\Rightarrow \widehat{A}\approx 53,{13}^0;$

Tam giác ABC có AB = AC nên tam giác ABC cân tại A

$\Rightarrow \widehat{B}=\widehat{C}=\frac{{180}^0-\widehat{A}}{2}\approx 63,{44}^0$.

Vậy $AB=AC=3\sqrt{5},BC=6,\widehat{A}=53,{13}^0,\widehat{B}=\widehat{C}=63,{44}^0.$

b) Gọi trực tâm H của tam giác ABC có tọa độ là H(x;y)

Khi đó, ta có: $\overrightarrow{AH}\left(x+4;y-1\right);\overrightarrow{BC}\left(0;-6\right);\overrightarrow{BH}\left(x-2;y-4\right);\overrightarrow{AC}\left(6;-3\right)$

Vì $AH\perp BC\Rightarrow \overrightarrow{AH}.\overrightarrow{BC}=0\iff \left(x+4\right).0+\left(y-1\right).\left(-6\right)=0\iff y=1.$

Vì $BH\perp AC\Rightarrow \overrightarrow{BH}.\overrightarrow{AC}=0\iff \left(x-2\right).6+\left(y-4\right).\left(-3\right)=0$

$$\begin{gathered} \iff \left(x-2\right).2+\left(y-4\right).\left(-1\right)=0 \\ \iff 2x-y=0 \end{gathered}$$

Mà y = 1 $\Rightarrow 2x-1=0\iff x=\frac{1}{2}.$

Ta có: $S_{ABC}=\frac{1}{2}AB.AC.\sin \left(\overrightarrow{AB},\overrightarrow{AC}\right)$

$=\frac{1}{2}AB.AC.\sqrt{1-co{\mathrm{s}}^2\left(\overrightarrow{AB},\overrightarrow{AC}\right)}$

$=\frac{1}{2}AB.AC.\sqrt{1-\frac{{\left(\overrightarrow{AB}.\overrightarrow{AC}\right)}^2}{{\overrightarrow{AB}}^2.{\overrightarrow{AC}}^2}}$

$=\frac{1}{2}AB.AC.\sqrt{\frac{{\overrightarrow{AB}}^2.{\overrightarrow{AC}}^2-{\left(\overrightarrow{AB}.\overrightarrow{AC}\right)}^2}{{\overrightarrow{AB}}^2.{\overrightarrow{AC}}^2}}$

$=\frac{1}{2}AB.AC.\frac{1}{AB.AC}.\sqrt{{\overrightarrow{AB}}^2.{\overrightarrow{AC}}^2-{\left(\overrightarrow{AB}.\overrightarrow{AC}\right)}^2}$

$=\frac{1}{2}\sqrt{{\overrightarrow{AB}}^2.{\overrightarrow{AC}}^2-{\left(\overrightarrow{AB}.\overrightarrow{AC}\right)}^2}$

$$\begin{gathered} {\mathrm{MB}}^2+{\mathrm{MC}}^2={\overrightarrow{\mathrm{MA}}}^2+{\overrightarrow{\mathrm{MB}}}^2+{\overrightarrow{\mathrm{MC}}}^2 \\ ={\left(\overrightarrow{\mathrm{MG}}+\overrightarrow{\mathrm{GA}}\right)}^2+{\left(\overrightarrow{\mathrm{MG}}+\overrightarrow{\mathrm{GB}}\right)}^2+{\left(\overrightarrow{\mathrm{MG}}+\overrightarrow{\mathrm{GC}}\right)}^2 \\ ={\overrightarrow{\mathrm{MG}}}^2+2\overrightarrow{\mathrm{MG}}.\overrightarrow{\mathrm{GA}}+{\overrightarrow{\mathrm{GA}}}^2+{\overrightarrow{\mathrm{MG}}}^2+2\overrightarrow{\mathrm{MG}}.\overrightarrow{\mathrm{GB}}+{\overrightarrow{\mathrm{GB}}}^2+{\overrightarrow{\mathrm{MG}}}^2+2\overrightarrow{\mathrm{MG}}.\overrightarrow{\mathrm{GC}}+{\overrightarrow{\mathrm{GC}}}^2 \\ =3{\overrightarrow{\mathrm{MG}}}^2+2\overrightarrow{\mathrm{MG}}.\left(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}\right)+{\overrightarrow{\mathrm{GA}}}^2+{\overrightarrow{\mathrm{GB}}}^2+{\overrightarrow{\mathrm{GC}}}^2 \\ \mathrm{Ta} \mathrm{có}: \overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=\overrightarrow{0}\mathrm{select}: (\mathrm{tính} \mathrm{chất} \mathrm{trọng} \mathrm{tâm} \mathrm{tam} \mathrm{giác}) \\ \Rightarrow \overrightarrow{\mathrm{MG}}.\left(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}\right)=\overrightarrow{\mathrm{MG}}.\overrightarrow{0}=0 \\ \Rightarrow {\mathrm{MA}}^2+{\mathrm{MB}}^2+{\mathrm{MC}}^2=3{\overrightarrow{\mathrm{MG}}}^2+{\overrightarrow{\mathrm{GA}}}^2+{\overrightarrow{\mathrm{GB}}}^2+{\overrightarrow{\mathrm{GC}}}^2. \end{gathered}$$