1. **Problem 7: Find the scalar product of two vectors A and B** given $A=5.0$, $B=8.0$, and the angle $\alpha=30^\circ$ between them. 2. The scalar product (dot product) formula is: $$\mathbf{A} \cdot \mathbf{B} = |A||B|\cos(\alpha)$$ where $|A|$ and $|B|$ are the magnitudes of vectors $\mathbf{A}$ and $\mathbf{B}$, and $\alpha$ is the angle between them. 3. Substitute the given values: $$5.0 \times 8.0 \times \cos(30^\circ)$$ 4. Recall that $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$. 5. Calculate the scalar product: $$5.0 \times 8.0 \times 0.866 = 40 \times 0.866 = 34.64$$ 6. **Answer for problem 7:** The scalar product is approximately $34.64$. --- 7. **Problem 8: Calculate the torque $\tau$ about the origin for a particle at position vector $\mathbf{r} = \mathbf{i} + \mathbf{j}$ m with force $\mathbf{F} = 2\mathbf{i} + 3\mathbf{j}$ N acting on it.** 8. Torque is given by the cross product: $$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$$ 9. Write vectors in component form: $$\mathbf{r} = (1,1,0), \quad \mathbf{F} = (2,3,0)$$ 10. Compute the cross product using the determinant formula: $$\boldsymbol{\tau} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 2 & 3 & 0 \end{vmatrix} = (1 \times 0 - 0 \times 3)\mathbf{i} - (1 \times 0 - 0 \times 2)\mathbf{j} + (1 \times 3 - 1 \times 2)\mathbf{k}$$ 11. Simplify: $$\boldsymbol{\tau} = 0\mathbf{i} - 0\mathbf{j} + (3 - 2)\mathbf{k} = \mathbf{k}$$ 12. The torque vector is: $$\boldsymbol{\tau} = \mathbf{k}$$ which means the torque has magnitude 1 N·m in the positive $z$-direction. 13. **Answer for problem 8:** The torque about the origin is $1$ N·m in the $\mathbf{k}$ direction (out of the plane). --- **Summary:** - Scalar product $= 34.64$ - Torque vector $= \mathbf{k}$ with magnitude $1$ N·m