For the two vectors in Fig. E1.35, find the magnitude and direction of (a) the vector product A x B

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1. Identify the magnitudes and directions of vectors \( \vec{A} \) and \( \vec{B} \). \( \vec{A} \) has a magnitude of 2.80 cm and is at an angle of 60° above the positive x-axis. \( \vec{B} \) has a magnitude of 1.90 cm and is at an angle of 60° below the negative x-axis.
2. Use the formula for the vector product (cross product) \( \vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin(\theta) \hat{n} \), where \( \theta \) is the angle between the vectors and \( \hat{n} \) is the unit vector perpendicular to the plane containing \( \vec{A} \) and \( \vec{B} \).
3. Determine the angle \( \theta \) between the vectors. Since \( \vec{A} \) is 60° above the x-axis and \( \vec{B} \) is 60° below the negative x-axis, the angle between them is 180° - 60° - 60° = 60°.
4. Calculate the magnitude of the cross product using the magnitudes of \( \vec{A} \) and \( \vec{B} \) and the sine of the angle between them: \( |\vec{A} \times \vec{B}| = (2.80 \text{ cm})(1.90 \text{ cm}) \sin(60°) \).
5. Determine the direction of the cross product using the right-hand rule. Point your fingers in the direction of \( \vec{A} \) and curl them towards \( \vec{B} \). Your thumb points in the direction of the cross product, which is out of the plane of the page (positive z-direction).