1. **Problem 1(a):** Given vectors $ \overrightarrow{PQ} = \begin{pmatrix} -3 \\ 7 \end{pmatrix} $ and $ 4\overrightarrow{PR} = \begin{pmatrix} -2 \\ 8 \end{pmatrix} $, find $ \overrightarrow{RQ} $. 2. **Step 1:** Find $ \overrightarrow{PR} $ by dividing $ 4\overrightarrow{PR} $ by 4: $$ \overrightarrow{PR} = \frac{1}{4} \begin{pmatrix} -2 \\ 8 \end{pmatrix} = \begin{pmatrix} \frac{-2}{4} \\ \frac{8}{4} \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} \\ 2 \end{pmatrix} $$ 3. **Step 2:** Use the vector addition rule: $$ \overrightarrow{RQ} = \overrightarrow{PQ} - \overrightarrow{PR} $$ 4. **Step 3:** Substitute the values: $$ \overrightarrow{RQ} = \begin{pmatrix} -3 \\ 7 \end{pmatrix} - \begin{pmatrix} -\frac{1}{2} \\ 2 \end{pmatrix} = \begin{pmatrix} -3 + \frac{1}{2} \\ 7 - 2 \end{pmatrix} = \begin{pmatrix} -\frac{6}{2} + \frac{1}{2} \\ 5 \end{pmatrix} = \begin{pmatrix} -\frac{5}{2} \\ 5 \end{pmatrix} $$ --- **Problem 1(b):** Given vectors $ \mathbf{a} = \alpha i + 6j $, $ \mathbf{b} = 4i + \beta j $, and $ \mathbf{c} = (2\alpha + 5\beta) i + 20j $, and the relation $ \mathbf{c} = 3\mathbf{a} - 2\mathbf{b} $, find $ \alpha $ and $ \beta $. 1. **Step 1:** Write the right side explicitly: $$ 3\mathbf{a} - 2\mathbf{b} = 3(\alpha i + 6j) - 2(4i + \beta j) = (3\alpha i + 18j) - (8i + 2\beta j) = (3\alpha - 8) i + (18 - 2\beta) j $$ 2. **Step 2:** Equate components of $ \mathbf{c} $ and $ 3\mathbf{a} - 2\mathbf{b} $: $$ 2\alpha + 5\beta = 3\alpha - 8 $$ $$ 20 = 18 - 2\beta $$ 3. **Step 3:** Solve the second equation for $ \beta $: $$ 20 = 18 - 2\beta \implies 20 - 18 = -2\beta \implies 2 = -2\beta \implies \beta = -1 $$ 4. **Step 4:** Substitute $ \beta = -1 $ into the first equation: $$ 2\alpha + 5(-1) = 3\alpha - 8 \implies 2\alpha - 5 = 3\alpha - 8 $$ 5. **Step 5:** Rearrange and simplify: $$ 2\alpha - 5 = 3\alpha - 8 \implies -5 + 8 = 3\alpha - 2\alpha \implies 3 = \alpha $$ **Final answers:** $$ \overrightarrow{RQ} = \begin{pmatrix} -\frac{5}{2} \\ 5 \end{pmatrix}, \quad \alpha = 3, \quad \beta = -1 $$