Question

4) Найдите косинус угла между векторами а(-3;2) и в(4;1).

Answer 1

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Answer 2

$\displaystyle \tt \cos\alpha =\frac{(\vec a;\vec b)}{|\vec a|\cdot| \vec b|}=\frac{a_x\cdot b_x+a_y\cdot b_y}{\sqrt{a_x^2+a_y^2}\cdot\sqrt{b_x^2} +b_y^2}\Longrightarrow \cos\alpha =\frac{(\vec a;\vec b)}{|\vec a|\cdot|\vec b|} =\frac{-3\cdot4+2\cdot1}{\sqrt{(-3)^2+2^2}\cdot \sqrt{4^2+1^2} }=\frac{-10}{\sqrt{(9+4)\cdot (16+1)} } =\frac{-10}{\sqrt{13\cdot17} } =\frac{-10\cdot\sqrt{221} }{\sqrt{221}\cdot\sqrt{221} } =\boxed{\tt \Huge -\frac{10\sqrt{221} }{221} } .$