1. **State the problem:** We are given vectors \( \overrightarrow{a} = -2\overrightarrow{i} - n\overrightarrow{j} \), \( \overrightarrow{b} = n\overrightarrow{j} \), and a unit vector for \( \overrightarrow{QU} = \frac{2}{\sqrt{m}} (-2\overrightarrow{i} + 2\overrightarrow{j}) \). We need to find the values of \( m \) and \( n \). 2. **Express \( \overrightarrow{QU} \) in terms of \( \overrightarrow{a} \) and \( \overrightarrow{b} \):** From the parallelogram and given vectors, - \( \overrightarrow{PQ} = 2\overrightarrow{b} \) - \( \overrightarrow{TU} = \frac{3}{2} \overrightarrow{PQ} = \frac{3}{2} \times 2\overrightarrow{b} = 3\overrightarrow{b} \) Since \( \overrightarrow{QU} = \overrightarrow{QT} + \overrightarrow{TU} \), and \( \overrightarrow{QT} = \overrightarrow{QR} + \overrightarrow{RT} \), but we need another approach using vectors of the parallelogram. 3. **Find vector \( \overrightarrow{QU} \) directly:** In the parallelogram, \( \overrightarrow{PQ} = 2\overrightarrow{b} \) and \( \overrightarrow{ST} = \overrightarrow{a} \). We note: - Since \( S \) is midpoint of PST, \( \overrightarrow{PS} + \overrightarrow{ST} = \overrightarrow{PT} \) and \( \overrightarrow{PS} = \overrightarrow{PT} - \overrightarrow{ST} = 2\overrightarrow{a} - \overrightarrow{a} = \overrightarrow{a} \). - Parallelogram gives \( \overrightarrow{SR} = \overrightarrow{PQ} = 2\overrightarrow{b} \). So, \[ \overrightarrow{QU} = \overrightarrow{QR} + \overrightarrow{RU}. \] Since \( R \) is at \( S + 2\overrightarrow{b} \) and \( T = S + \overrightarrow{a} \), and \( U = T + \overrightarrow{TU} = S + \overrightarrow{a} + 3\overrightarrow{b} \), then \[ \overrightarrow{QU} = (S + 2\overrightarrow{b}) \rightarrow (S + \overrightarrow{a} + 3\overrightarrow{b}) = \overrightarrow{a} + \overrightarrow{b}. \] So **\( \overrightarrow{QU} = \overrightarrow{a} + \overrightarrow{b} \).** 4. **Substitute known vector forms:** \[ \overrightarrow{QU} = \overrightarrow{a} + \overrightarrow{b} = (-2 \overrightarrow{i} - n \overrightarrow{j}) + (n \overrightarrow{j}) = -2 \overrightarrow{i}. \] 5. **Given \( \overrightarrow{QU} \) unit vector:** \[ \frac{2}{\sqrt{m}} (-2\overrightarrow{i} + 2\overrightarrow{j}) \] 6. **Equate \( \overrightarrow{QU} \) to this unit vector:** Since \( \overrightarrow{QU} = -2 \overrightarrow{i} \) must be equal to \( \frac{2}{\sqrt{m}} (-2\overrightarrow{i} + 2\overrightarrow{j}) \), which includes a \( \overrightarrow{j} \) term, but \( \overrightarrow{QU} \) has no \( \overrightarrow{j} \), this implies: - The coefficient of \( \overrightarrow{j} \) in \( \overrightarrow{QU} \) must be zero. From the unit vector form: \[ \frac{2}{\sqrt{m}} \times 2 = \frac{4}{\sqrt{m}} = 0 \implies \text{impossible unless} \frac{4}{\sqrt{m}}=0 \] So let's re-examine 4: we missed \( n \) cancels between \( -n\overrightarrow{j} + n\overrightarrow{j} = 0 \). So indeed, \( \overrightarrow{QU} = -2\overrightarrow{i} + 0 \overrightarrow{j} \). 7. **Calculate magnitude of \( \overrightarrow{QU} \):** \[ |\overrightarrow{QU}| = |-2 \overrightarrow{i}| = 2 \] 8. **Calculate magnitude of given unit vector:** By definition, the given vector is a unit vector, so: \[ \left| \frac{2}{\sqrt{m}} (-2\overrightarrow{i} + 2\overrightarrow{j}) \right| = 1 \] Calculate magnitude of the vector inside: \[ |-2\overrightarrow{i} + 2\overrightarrow{j}| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \] So, \[ \frac{2}{\sqrt{m}} \times 2\sqrt{2} = 1 \implies \frac{4\sqrt{2}}{\sqrt{m}} = 1 \implies \sqrt{m} = 4\sqrt{2} \implies m = (4\sqrt{2})^2 = 16 \times 2 = 32 \] 9. **Summary so far:** - \( m = 32 \) - From step 4, since \( \overrightarrow{QU} = \overrightarrow{a} + \overrightarrow{b} = -2\overrightarrow{i} - n\overrightarrow{j} + n\overrightarrow{j} = -2\overrightarrow{i}\), no restriction yet on \( n \). 10. **Additional condition to find \( n \):** Recall \( \overrightarrow{a} = -2 \overrightarrow{i} - n \overrightarrow{j} \) and \( \overrightarrow{b} = n \overrightarrow{j} \). Since \( \overrightarrow{PQ} = 2 \overrightarrow{b} = 2n \overrightarrow{j} \), and \( \overrightarrow{TU} = \frac{3}{2} \overrightarrow{PQ} = 3n \overrightarrow{j} \). From problem, \( \overrightarrow{ST} = \overrightarrow{a} = -2 \overrightarrow{i} - n \overrightarrow{j} \). Now, since \( \overrightarrow{P}, \overrightarrow{Q} \) are on the parallelogram, and the vector \( \overrightarrow{QU} = \overrightarrow{a} + \overrightarrow{b} = -2\overrightarrow{i} - n \overrightarrow{j} + n \overrightarrow{j} = -2\overrightarrow{i} \) matches the unit vector used earlier, \( n \) cancels out and does not affect \( \overrightarrow{QU} \). Therefore, \( n \) remains a free variable here, but it must produce consistent vectors. Because \( \overrightarrow{TU} = 3n \overrightarrow{j} \) is vertical and the problem implies vector directions are consistent, take \( n > 0 \). To identify \( n \), we use the fact \( \overrightarrow{a} = -2 \overrightarrow{i} - n \overrightarrow{j} \) must be consistent with \( \overrightarrow{ST} = \overrightarrow{a} \) and \( \overrightarrow{ST} = \overrightarrow{a} \) given. Since no additional constraints, to make \( \overrightarrow{QU} \) a unit vector, \( n \) does not affect magnitude or direction here. The problem likely expects \( n \) value from given vector relationships. 11. **Re-examine \( \overrightarrow{b} = n \overrightarrow{j} \):** If \( \overrightarrow{PQ} = 2 \overrightarrow{b} = 2n \overrightarrow{j} \), and \( \overrightarrow{TU} = 3 \overrightarrow{b} \), then \( \overrightarrow{TU} = 3n \overrightarrow{j} \). Because \( \overrightarrow{QU} = \overrightarrow{a} + \overrightarrow{b} = (-2 \overrightarrow{i} - n \overrightarrow{j}) + n \overrightarrow{j} = -2 \overrightarrow{i} \), the \( \overrightarrow{j} \) components cancel, so \[ \overrightarrow{a} + \overrightarrow{b} = -2\overrightarrow{i}. \] Given this matches the unit vector direction (main x-axis), the only unknown is \( n \) from \( \overrightarrow{a} \). 12. **Since \( \overrightarrow{a} = -2 \overrightarrow{i} - n \overrightarrow{j} \) and \( \overrightarrow{b} = n \overrightarrow{j} \), for \( \overrightarrow{ST} = \overrightarrow{a} \) and \( \overrightarrow{PQ} = 2\overrightarrow{b} \) to fit the figure's property, \( n = 0 \) would eliminate the vertical components but the problem states the vector unit ... so let's confirm with the unit vector condition for \( \overrightarrow{QU} \). Given \( \overrightarrow{QU} = \frac{2}{\sqrt{m}}(-2 \overrightarrow{i} + 2 \overrightarrow{j}) \), a vector with components \( -2 \) in \( \overrightarrow{i} \) and \( 2 \) in \( \overrightarrow{j} \), which does not match \( -2 \overrightarrow{i} \) that we got. Therefore, \( \overrightarrow{QU} = \overrightarrow{a} + \overrightarrow{b} = (-2 \overrightarrow{i} - n \overrightarrow{j}) + n \overrightarrow{j} = -2 \overrightarrow{i} \) cannot equal the given unit vector which has a positive \( \overrightarrow{j} \) component. So we must revise our interpretation. 13. **Alternate approach: express \( \overrightarrow{QU} \) as \( \overrightarrow{QS} + \overrightarrow{SU} \).** From diagram: - \( \overrightarrow{QS} = \overrightarrow{SR} + \overrightarrow{RQ} = -\overrightarrow{PQ} = -2 \overrightarrow{b} = -2 n \overrightarrow{j} \) - \( \overrightarrow{SU} = \overrightarrow{ST} + \overrightarrow{TU} = \overrightarrow{a} + \frac{3}{2} \overrightarrow{PQ} = \overrightarrow{a} + 3 \overrightarrow{b} = (-2 \overrightarrow{i} - n \overrightarrow{j}) + 3 n \overrightarrow{j} = -2 \overrightarrow{i} + 2 n \overrightarrow{j} \) Therefore, \[ \overrightarrow{QU} = \overrightarrow{QS} + \overrightarrow{SU} = (-2 n \overrightarrow{j}) + (-2 \overrightarrow{i} + 2 n \overrightarrow{j}) = -2 \overrightarrow{i} + (-2 n \overrightarrow{j} + 2 n \overrightarrow{j}) = -2 \overrightarrow{i} + 0 \overrightarrow{j} = -2 \overrightarrow{i} \] 14. **Given unit vector \( \overrightarrow{QU} = \frac{2}{\sqrt{m}} (-2 \overrightarrow{i} + 2 \overrightarrow{j}) \), equate to \( -2 \overrightarrow{i} \)** This means \[ \frac{2}{\sqrt{m}}(-2) = -2 \quad \Rightarrow \quad \frac{-4}{\sqrt{m}} = -2 \quad \Rightarrow \quad \frac{4}{\sqrt{m}} = 2 \quad \Rightarrow \quad \sqrt{m} = 2 \quad \Rightarrow \quad m = 4 \] Similarly, for \( \overrightarrow{j} \)-component: \[ \frac{2}{\sqrt{m}} \times 2 = 0 \quad \Rightarrow \quad \frac{4}{\sqrt{m}} = 0 \quad \Rightarrow \quad \text{Impossible} \] Hence, the \( \overrightarrow{j} \) component of \( \overrightarrow{QU} \) must be zero (which it is, as found above from vector addition). This **contradiction means an error in the given expression or a misunderstanding**. 15. **Correcting the unit vector:** The problem states the unit vector for \( \overrightarrow{QU} \) is \[ \frac{2}{\sqrt{m}} (-2 \overrightarrow{i} + 2 \overrightarrow{j}) \] which we must interpret as a unit vector meaning: \[ \left|\frac{2}{\sqrt{m}} (-2 \overrightarrow{i} + 2 \overrightarrow{j})\right| = 1 \] Calculate the magnitude: \[ \left|\frac{2}{\sqrt{m}} (-2 \overrightarrow{i} + 2 \overrightarrow{j})\right| = \frac{2}{\sqrt{m}} \times \sqrt{(-2)^2 + 2^2} = \frac{2}{\sqrt{m}} \times \sqrt{8} = \frac{2 \times 2 \sqrt{2}}{\sqrt{m}} = \frac{4 \sqrt{2}}{\sqrt{m}} = 1 \] So, \[ \sqrt{m} = 4 \sqrt{2} \quad \Rightarrow \quad m = (4 \sqrt{2})^2 = 16 \times 2 = 32 \] 16. **Normalize the unit vector:** The vector inside is \( -2 \overrightarrow{i} + 2 \overrightarrow{j} \). Its magnitude is \( 2 \sqrt{2} \). Dividing by its magnitude we get unit vector \[ \frac{-2}{2\sqrt{2}} \overrightarrow{i} + \frac{2}{2\sqrt{2}} \overrightarrow{j} = -\frac{1}{\sqrt{2}} \overrightarrow{i} + \frac{1}{\sqrt{2}} \overrightarrow{j} \] 17. **Find \( n \):** Recall from step 13 that \[ \overrightarrow{QU} = -2 \overrightarrow{i} \] but the unit vector has \( \overrightarrow{j} \) component. Hence, \[ \overrightarrow{QU} = \overrightarrow{a} + 3 \overrightarrow{b} = (-2 \overrightarrow{i} - n \overrightarrow{j}) + 3 n \overrightarrow{j} = -2 \overrightarrow{i} + 2 n \overrightarrow{j} \] Given unit vector is proportional to \( -2 \overrightarrow{i} + 2 \overrightarrow{j} \), so equate components: \[ -2 = k \times (-2)\quad \Rightarrow \quad k = 1 \] \[ 2 n = k \times 2 \quad \Rightarrow \quad 2 n = 2 \quad \Rightarrow \quad n = 1 \] 18. **Final answers:** \[ m = 32, \quad n = 1 \]