Calculate correct to one decimal place, the angle between 5i + 12j and -2i + 3J
Cos\(\theta\) = \(\frac{ a . b}{|a| |b|}\)
a.b = (5 x - 2) + (12 x 3) = -10 + 36 = 26
|a| = \(\sqrt{ 5^2 + 12^2}\) = \(\sqrt{25 + 144}\) = \(\sqrt{169}\) = 13
|b| = \(\sqrt{ -2^2 + 3^2}\) = \(\sqrt{4 + 9}\) = \(\sqrt{13}\) ≈ 3.606
Cos\(\theta\) = \(\frac{26}{13\sqrt{13}}\) = \(\frac{2}{\sqrt{13}}\)
\(\theta\) = cos\(^{-1}\)(\(\frac{2}{\sqrt{13}}\)) ≈ 56.3099º
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