Question

Mulțumesc anticipat!​

Answer 1

Salut,

$\vec{a}=a_x\cdot\vec{i}+a_y\cdot\vec{j}=6\cdot\vec{i}+3\cdot\vec{j}\\\\\vec{b}=a=b_x\cdot\vec{i}-4\cdot\vec{j}.\\\\Produsul\ scalar\ al\ vectorilor\ \vec{a},\ \vec{b}\ este:\\\\\vec{a}\cdot\vec{b}=|\vec{a}|\cdot|\vec{b}|\cdot\cos\angle{(\vec{a},\vec{b}}).\\\\Vectorii\ sunt\ ortogonali\ (perpendiculari)\ dac\breve{a}\ \cos\angle{(\vec{a},\vec{b}})=0,\ deci:\\\\\cos\angle{(\vec{a},\vec{b}})=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\cdot|\vec{b}|}=\dfrac{(6\cdot\vec{i}+3\cdot\vec{j})(b_x\cdot\vec{i}-4\cdot\vec{j})}{|\vec{a}|\cdot|\vec{b}|}=\dfrac{6\cdot b_x+3\cdot(-4)}{|\vec{a}|\cdot|\vec{b}|}=\\\\=\dfrac{6\cdot b_x-12}{|\vec{a}|\cdot|\vec{b}|}.\ Deci\ 6\cdot b_x-12=0\Rightarrow 6\cdot b_x=12,\ deci\ b_x=2.$

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