If \(x = i - 3j\) and \(y = 6i + j\), calculate the angle between x and y.
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Correct Answer: Option C
Explanation:
\(\overrightarrow{x} . \overrightarrow{y} = |\overrightarrow{x}||\overrightarrow{y}|\cos\theta\)
\(\overrightarrow{x} . \overrightarrow{y} = (i - 3j) . (6i + j) = 6 - 3 = 3\)
\(|\overrightarrow{x}| = \sqrt{1^{2} + (-3)^{2}} = \sqrt{10}\)
\(|\overrightarrow{y}| = \sqrt{6^{2} + 1^{2}} = \sqrt{37}\)
\(\therefore 3 = (\sqrt{10})(\sqrt{37})\cos \theta\)
\(\cos\theta = \frac{3}{\sqrt{370}} = 0.1559\)
\(\theta = \cos^{-1} 0.1559 \approxeq 81°\)
\(\overrightarrow{x} . \overrightarrow{y} = |\overrightarrow{x}||\overrightarrow{y}|\cos\theta\)
\(\overrightarrow{x} . \overrightarrow{y} = (i - 3j) . (6i + j) = 6 - 3 = 3\)
\(|\overrightarrow{x}| = \sqrt{1^{2} + (-3)^{2}} = \sqrt{10}\)
\(|\overrightarrow{y}| = \sqrt{6^{2} + 1^{2}} = \sqrt{37}\)
\(\therefore 3 = (\sqrt{10})(\sqrt{37})\cos \theta\)
\(\cos\theta = \frac{3}{\sqrt{370}} = 0.1559\)
\(\theta = \cos^{-1} 0.1559 \approxeq 81°\)