Qual o menor angulo entre os vetores?

\[\theta = \arccos({u_iv_i+u_jv_j+u_kv_k \over |\overrightarrow{u}| \cdot |\overrightarrow{v}|})\]

Onde \(|\overrightarrow{u}|=\sqrt{u_i^2+ u_j^2 +u_k^2}\)e \(|\overrightarrow{v}|=\sqrt{v_i^2+ v_j^2 +v_k^2}\)

Portanto, o ângulo entre os vetores \(\overrightarrow{a}=3 \hat{i}-4 \hat{j}\)e \(\overrightarrow{b}=-2 \hat{i}+3 \hat{k}\)é:

\[\eqalign{ \theta_{ab} &= \arccos({a_ib_i+a_jb_j+a_kb_k \over |\overrightarrow{a}| \cdot |\overrightarrow{b}|}) \\ &= \arccos({a_ib_i+a_jb_j+a_kb_k \over \sqrt{a_i^2+ a_j^2 +a_k^2} \cdot \sqrt{b_i^2+ b_j^2 +b_k^2}}) \\ &= \arccos({3\cdot(-2)+(-4)\cdot 0+0\cdot 3 \over \sqrt{3^2+ (-4)^2 +0^2} \cdot \sqrt{(-2)^2+ 0^2 +3^2}}) \\ &= \arccos({-6+0+0\over \sqrt{9+16} \cdot \sqrt{4+9}}) \\ &= \arccos({-6\over \sqrt{25} \cdot \sqrt{13}}) \\ &= \arccos({-6\over 5 \cdot \sqrt{13}}) \\ &=109,44^{\circ} }\]

Resposta correta: segunda alternativa - \(109,44^{\circ}\)