1. **Problem 1: Find p and q given vector $\mathbf{a} = p\mathbf{i} + q\mathbf{j}$ with $|\mathbf{a}|=11$ and angle 20° with positive y-axis.** 2. The magnitude of $\mathbf{a}$ is given by: $$|\mathbf{a}| = \sqrt{p^2 + q^2} = 11$$ 3. The angle $\theta$ between $\mathbf{a}$ and the positive y-axis is 20°. Since $\mathbf{a} = p\mathbf{i} + q\mathbf{j}$, the angle with y-axis satisfies: $$\cos 20^\circ = \frac{\mathbf{a} \cdot \mathbf{j}}{|\mathbf{a}||\mathbf{j}|} = \frac{q}{11}$$ 4. Solve for $q$: $$q = 11 \cos 20^\circ \approx 11 \times 0.9397 = 10.337$$ 5. Use magnitude to find $p$: $$p^2 + q^2 = 11^2 = 121$$ $$p^2 = 121 - (10.337)^2 = 121 - 106.87 = 14.13$$ $$p = \sqrt{14.13} \approx 3.76$$ --- 6. **Problem 2a: Prove $\overrightarrow{PQ}$ is parallel to $\overrightarrow{OB}$.** 7. Let $O=(0,0)$, $A=(0,a)$, $B=(b,0)$ for some positive $a,b$. 8. Point $P$ divides $OA$ in ratio $1:k$ from $O$ to $A$: $$P = \frac{k}{k+1} O + \frac{1}{k+1} A = \left(0, \frac{a}{k+1}\right)$$ 9. Point $Q$ divides $BA$ in ratio $1:k$ from $B$ to $A$: $$Q = \frac{k}{k+1} B + \frac{1}{k+1} A = \left(\frac{kb}{k+1}, \frac{a}{k+1}\right)$$ 10. Vector $\overrightarrow{PQ} = Q - P = \left(\frac{kb}{k+1} - 0, \frac{a}{k+1} - \frac{a}{k+1}\right) = \left(\frac{kb}{k+1}, 0\right)$ 11. Vector $\overrightarrow{OB} = B - O = (b,0)$ 12. Since $\overrightarrow{PQ} = \frac{k}{k+1} \overrightarrow{OB}$, $\overrightarrow{PQ}$ is parallel to $\overrightarrow{OB}$. --- 13. **Problem 2b: Given $|\overrightarrow{PQ}| = \frac{1}{3} |\overrightarrow{OB}|$, find $k$.** 14. Magnitudes: $$|\overrightarrow{PQ}| = \left|\frac{kb}{k+1}, 0\right| = \frac{kb}{k+1}$$ $$|\overrightarrow{OB}| = |(b,0)| = b$$ 15. Given: $$\frac{kb}{k+1} = \frac{1}{3} b$$ 16. Cancel $b$ (nonzero): $$\frac{k}{k+1} = \frac{1}{3}$$ 17. Cross multiply: $$3k = k + 1$$ $$3k - k = 1$$ $$2k = 1$$ $$k = \frac{1}{2}$$ **Final answers:** - $p \approx 3.76$ - $q \approx 10.34$ - $\overrightarrow{PQ}$ is parallel to $\overrightarrow{OB}$ - $k = \frac{1}{2}$