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What is the angle between the vectors A= 3i + 3j + 3k B= 2i + 1j + 3k

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Answer to a math question What is the angle between the vectors A= 3i + 3j + 3k B= 2i + 1j + 3k

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Neal
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A=3i+3j+3k B=2i+1j+3k A\cdot B=\esquerda(3\vezes2\direita)+\esquerda(3\vezes1\direita)+\esquerda(3\vezes3\direita)=6+3+9=18 \esquerda|A\direita|=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3} \esquerda|B\direita|=\sqrt{2^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14} Deixe,\:o\:ângulo\:entre\:dois\:vetores=\theta \cos\theta=\frac{A\cdot B}{\left|A\right|\left|B\right|}=\frac{18}{3\sqrt{3}\times\sqrt{14 }}=\frac{\sqrt{42}}{7} \theta=\cos^{-1}\left(\frac{\sqrt{42}}{7}\right)=22,21^{\circ} Ângulo\:entre\:dois\:vetores\:=22,21^{\circ}

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