Cho hai vec-tơ  $(\overrightarrow{s}+2 \overrightarrow{t})$  và  $(5 \overrightarrow{s}-4 \overrightarrow{t}  )$  vuông góc với nhau.
Xác định góc  $(\overrightarrow{s}, \overrightarrow{t}  )$,  biết rằng  $|\overrightarrow{s} |=|\overrightarrow{t} |=1$.
Vì  $(\overrightarrow{s}+2 \overrightarrow{t}  )$  và  $(5 \overrightarrow{s}-4 \overrightarrow{t}  )$  vuông góc với nhau
$\Rightarrow    (\overrightarrow{s}+2 \overrightarrow{t}  )(5 \overrightarrow{s}-4 \overrightarrow{t}  )=0$
$\Rightarrow    5 \overrightarrow{s}^2-4 \overrightarrow{s}.\overrightarrow{t}+10 \overrightarrow{t}.\overrightarrow{s}-8 \overrightarrow{t}^2=0      $
$\Rightarrow    \overrightarrow{s}.\overrightarrow{t}=\frac{1}{2}   $  vì  $\overrightarrow{s}^2=|\overrightarrow{s} |^2=1;   \overrightarrow{t}^2=|\overrightarrow{t} |^2=1  $
mà    $\overrightarrow{s}.\overrightarrow{t}=|\overrightarrow{t} |.|\overrightarrow{s} |.\cos (\overrightarrow{s},\overrightarrow{t}  )=\cos(\overrightarrow{s},\overrightarrow{t}  )  $
nên:   $\cos(\overrightarrow{s},\overrightarrow{t}  )=\frac{1}{2}    \Rightarrow     (\overrightarrow{s},\overrightarrow{t}  )=60^\circ$.

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