Cho hai vectơ i,j vuông góc, cùng có độ dài bằng 1. a) Tính

* Hướng dẫn giải

Hai vectơ $\overrightarrow{i},\overrightarrow{j}$vuông góc nên $(\overrightarrow{i},\overrightarrow{j})=90°$ .

Ta có: $\overrightarrow{i}.\overrightarrow{j}=\left|\overrightarrow{i}\right|.\left|\overrightarrow{j}\right|.cos(\overrightarrow{i},\overrightarrow{j})=\mathrm{1.1.}cos90°=0$ .

a) ${(\overrightarrow{i}+\overrightarrow{j})}^2={\overrightarrow{i}}^2+2\overrightarrow{i}.\overrightarrow{j}+{\overrightarrow{j}}^2$$={\left|\overrightarrow{i}\right|}^2+2\overrightarrow{i}.\overrightarrow{j}+{\left|\overrightarrow{j}\right|}^2$ = 1² + 2 . 0 + 1² = 2.

   ${(\overrightarrow{i}-\overrightarrow{j})}^2={\overrightarrow{i}}^2-2\overrightarrow{i}.\overrightarrow{j}+{\overrightarrow{j}}^2$$={\left|\overrightarrow{i}\right|}^2-2\overrightarrow{i}.\overrightarrow{j}+{\left|\overrightarrow{j}\right|}^2$ = 1² – 2 . 0 + 1² = 2.

$(\overrightarrow{i}+\overrightarrow{j}).(\overrightarrow{i}-\overrightarrow{j})={\overrightarrow{i}}^2-{\overrightarrow{j}}^2={\left|\overrightarrow{i}\right|}^2-{\left|\overrightarrow{j}\right|}^2=1^2-1^2=0$

b) Ta có: $\overrightarrow{a}.\overrightarrow{b}=(2\overrightarrow{i}+2\overrightarrow{j}).(3\overrightarrow{i}-3\overrightarrow{j})$$=2(\overrightarrow{i}+\overrightarrow{j}).3(\overrightarrow{i}-\overrightarrow{j})=6.(\overrightarrow{i}+\overrightarrow{j}).(\overrightarrow{i}-\overrightarrow{j})=6.0=0$ .

Do đó $\overrightarrow{a}.\overrightarrow{b}=0\iff \overrightarrow{a}\perp \overrightarrow{b}$ .

Vậy $(\overrightarrow{a},\overrightarrow{b})=90°$ .