Cho hai vecto cùng phương $\overrightarrow{u}=\left(x;y\right)$ và $\overrightarrow{v}=\left(kx;ky\right).$ Hãy kiểm tra công thức $\overrightarrow{u}.\overrightarrow{v}=k\left(x^2+y^2\right)$ theo từng trường hợp sau:

a) $\overrightarrow{u}=\overrightarrow{0};$

b) $\overrightarrow{u}\ne \overrightarrow{0}$ và $k\ge 0;$

c) $\overrightarrow{u}\ne \overrightarrow{0}$ và k < 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 $\Rightarrow 2x-1=0\iff x=\frac{1}{2}.$

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

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

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