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If vectors \(P\), \(Q\), and \(R\) have magnitudes 5, 12, and 13 units respectively and \(P+Q=R\), then the angle between \(Q\) and \(R\) is

1. \(\cos^{-1}\!\left(\frac{5}{12}\right)\)
2. \(\cos^{-1}\!\left(\frac{5}{13}\right)\)
3. \(\cos^{-1}\!\left(\frac{12}{13}\right)\)
4. \(\cos^{-1}\!\left(\frac{2}{13}\right)\)

Correct answer: \(\cos^{-1}\!\left(\frac{12}{13}\right)\)

Solution

From \(R=P+Q\), we have \(|R|^2=|P|^2+|Q|^2+2|P||Q|\cos\theta\), where \(\theta\) is the angle between \(P\) and \(Q\). Using 13, 5, and 12 gives \(169=25+144+120\cos\theta\), so \(\cos\theta=0\), meaning \(P\perp Q\). Then the angle between \(Q\) and \(R\) satisfies \(\cos\phi=\frac{Q\cdot R}{|Q||R|}=\frac{Q\cdot(P+Q)}{12\cdot 13}=\frac{144}{156}=\frac{12}{13}\).

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