The angle between the two vectors is 117⁰
The given parameters;
vector A = 3.00i + 1.00j
vector B = -3.00i + 3.00j
The angle between the two vectors is calculated as follows;
[tex]cos \ \theta = \frac{A\ . \ B}{|A| \ . \ |B|}[/tex]
The dot product of vector A and B is calculated as;
[tex]A \ . \ B = (3i \ + j) \ . \ (-3i \ + 3j) = (3\times -3) + (1 \times 3) = -9 + 3 =- 6[/tex]
The magnitude of vector A and B is calculated as;
[tex]|A| = \sqrt{3^2 + 1^2} = \sqrt{10} \\\\|B| = \sqrt{(-3)^2 + (3)^2} = \sqrt{18}[/tex]
The angle between the two vectors is calculated as;
[tex]cos \ \theta = \frac{-6}{\sqrt{10} \ . \sqrt{18} } \\\\cos \ \theta = \frac{-6}{\sqrt{180} } \\\\cos \ \theta = -0.4472\\\\\theta = cos \ ^{-1} (-0.4472) \\\\\theta = 116.6^0 \approx 117^0[/tex]
Thus, the angle between the two vectors is 117⁰
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