Question

Determinati, in fiecare caz, unghiul vectorilor:
punctele a,b,f​

Answer 1

Salut! ⚛

_________

 Determinați, în fiecare caz, unghiul vectorilor.

Fie u(x₁, y₁) și v(x₂, y₂) și ∝ = m(∡u, v)

$a)$ $u(2, -1)$$,$ $v(1,2)$

$cos\alpha =\frac{x_{1} *x_{2}+y_{1} *y_{2} }{\sqrt{x_{1}^{2} +y_{1} ^{2} }*\sqrt{x_{2}^{2} +y_{2}^{2} } }$

$cos\alpha =\frac{2*1+(-1)*2}{\sqrt{4+1}*\sqrt{1+4} } => cos\alpha =\frac{2-2}{\sqrt{5} *\sqrt{5} } =\frac{0}{5} =0$

$cos\alpha =0 => \alpha =90 ^{0}$

$b)$ $u(4, -2),$ $v(1,2)$

$cos\alpha =\frac{x_{1} *x_{2}+y_{1} *y_{2} }{\sqrt{x_{1}^{2} +y_{1} ^{2} }*\sqrt{x_{2}^{2} +y_{2}^{2} } }$

$cos\alpha =\frac{4*1+(-2)*2}{\sqrt{16+4} *\sqrt{1+4} } =\frac{4-4}{\sqrt{20}*\sqrt{5} } =\frac{0}{\sqrt{100} } =0$

$cos\alpha =0 => \alpha =90 ^{0}$

$f)$ $u(4, -3),$ $v(-3, 4)$

$cos\alpha =\frac{x_{1} *x_{2}+y_{1} *y_{2} }{\sqrt{x_{1}^{2} +y_{1} ^{2} }*\sqrt{x_{2}^{2} +y_{2}^{2} } }$

$cos\alpha =\frac{4*(-3)+(-3)*4}{\sqrt{16+9} *\sqrt{9+16} } =\frac{-12-12}{\sqrt{25} *\sqrt{25} } =-\frac{24}{25}$

$cos(2\pi -\alpha )=-\frac{24}{25}=> 2\pi -\alpha =arccos(-\frac{24}{25} ) => 2\pi -\alpha =\pi -arccos(\frac{24}{25} )$    ${\mathbf{(1)}$

$cos(\alpha )=-\frac{24}{25} => \alpha =\pi -arccos(\frac{24}{25} )$   ${\mathbf{(2)}$

$Din$ $(1)$ $si$ $(2)$ $=>$ $2\pi -\alpha =\alpha => 2\alpha =2\pi => \alpha =\pi => \alpha =180^{0}$

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