Question

Cho tam giác ABC vuông tại A có \(\widehat{B}\)=60độ, BC=2cm. Tính \(\left|\overrightarrow{AB}\right|,\left|\overrightarrow{AC}\right|,\left|\overrightarrow{AB}+\overrightarrow{AC}\right|,\left|\overrightarrow{AB}-\overrightarrow{AC}\right|?\)

Answer 1

Lời giải:

\(|\overrightarrow{AB}|=BC\cos B=2.\cos 60^0=1\) (cm)

\(|\overrightarrow{AC}|=BC\sin B=2.\sin 60^0=\sqrt{3}\) (cm)

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Do tam giác $ABC$ vuông tại $A$ nên $\overrightarrow{AB}\perp \overrightarrow{AC}\Rightarrow \overrightarrow{AB}.\overrightarrow{AC}=0$. Do đó:

\(|\overrightarrow{AB}+\overrightarrow{AC}|^2=(\overrightarrow{AB}+\overrightarrow{AC})^2=AB^2+AC^2+2\overrightarrow{AB}.\overrightarrow{AC}\)

\(=BC^2+0=BC^2=4\) (cm)

$\Rightarrow |\overrightarrow{AB}+\overrightarrow{AC}|=2$ (cm)

Tương tự:

\(|\overrightarrow{AB}-\overrightarrow{AC}|^2=AB^2+AC^2-2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2=BC^2=4\)

$\Rightarrow |\overrightarrow{AB}-\overrightarrow{AC}|=2$ (cm)

Answer 2

Lời giải:

\(|\overrightarrow{AB}|=BC\cos B=2.\cos 60^0=1\) (cm)

\(|\overrightarrow{AC}|=BC\sin B=2.\sin 60^0=\sqrt{3}\) (cm)

------------------

Do tam giác $ABC$ vuông tại $A$ nên $\overrightarrow{AB}\perp \overrightarrow{AC}\Rightarrow \overrightarrow{AB}.\overrightarrow{AC}=0$. Do đó:

\(|\overrightarrow{AB}+\overrightarrow{AC}|^2=(\overrightarrow{AB}+\overrightarrow{AC})^2=AB^2+AC^2+2\overrightarrow{AB}.\overrightarrow{AC}\)

\(=BC^2+0=BC^2=4\) (cm)

$\Rightarrow |\overrightarrow{AB}+\overrightarrow{AC}|=2$ (cm)

Tương tự:

\(|\overrightarrow{AB}-\overrightarrow{AC}|^2=AB^2+AC^2-2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2=BC^2=4\)

$\Rightarrow |\overrightarrow{AB}-\overrightarrow{AC}|=2$ (cm)