1. **Problem Statement:** We analyze vector operations for $\mathbf{d} = \mathbf{i} - \mathbf{j} + \mathbf{k}$ and $\mathbf{b} = 2\mathbf{j} - 3\mathbf{k}$, verify given expressions, simplify algebraic and trigonometric identities, solve quadratic equations, and interpret graph specifications. 2. **Vector analysis:** - Compute the parallelogram spanned by $\mathbf{d}$ and $\mathbf{b}$. - Calculate the cross product $\mathbf{d} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 0 & 2 & -3 \end{vmatrix} = \mathbf{i}((-1)(-3)-1 \cdot 2) - \mathbf{j}(1 \cdot (-3) - 1 \cdot 0) + \mathbf{k}(1 \cdot 2 - (-1) \cdot 0) = \mathbf{i}(3-2) - \mathbf{j}(-3 - 0) + \mathbf{k}(2 - 0) = \mathbf{i}(1) + \mathbf{j}(3) + \mathbf{k}(2)$ 3. **Given determinant:** $\begin{vmatrix} 1+2i \\ 2-1 \end{vmatrix} = 1$ seems incomplete; likely referencing simple determinant or complex norm, but no further info stated. 4. **Algebraic expressions:** - Simplify $2x^2 + 3y^2$ is left as is since base form. 5. **Trigonometric identity:** - Prove $\sin\theta + \cos\theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)$. Using angle addition: $$\sin\theta + \cos\theta = \sqrt{2} \left(\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta \right) = \sqrt{2} \sin\left(\theta + \frac{\pi}{4} \right)$$ 6. **Simplify:** - $\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} = \sec \theta - \tan \theta$. Start lhs: $$\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} = \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}} = \frac{1 - \sin \theta}{|\cos \theta|}$$ Since $\sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta}$. Thus, equality holds assuming $\cos \theta > 0$. 7. **Simplify:** $\frac{1}{x^2 (x+1)}$ remains. 8. **Expressions with exponents:** - $(4 + 3x)^{5/2}$ remains as is. - $(2x^2 - \frac{3}{2x})^{10}$ remains as expression. 9. **Quadratic equation:** $x^2 - (3K + 1)x + 4x = 0$ simplifies to $x^2 - (3K - 3)x = 0$. 10. **Solve equation:** $\sqrt{2}x^2 - \sqrt{3}x + \sqrt{2} = 0$ - Using quadratic formula: $$x = \frac{\sqrt{3} \pm \sqrt{ (-\sqrt{3})^2 - 4\sqrt{2} \cdot \sqrt{2} }}{2\sqrt{2}} = \frac{\sqrt{3} \pm \sqrt{3 - 8}}{2\sqrt{2}} = \frac{\sqrt{3} \pm i\sqrt{5}}{2\sqrt{2}}$$ 11. **Solve:** $x^2 - 4x + 2 = 0$ - Using quadratic formula: $$x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}$$ 12. **Expand squares:** - $(a - b)^2 = a^2 - 2ab + b^2$ - $(a + b)^2 = a^2 + 2ab + b^2$ 13. **Quadratic with parameter $k$:** $4x^2 - 3x + k = 0$ - Roots depend on $k$; discriminant $\Delta = 9 - 16k$. 14. **Powers:** - $x^4$ straightforward. - $(2x + \frac{1}{x})^{8}$ indicates binomial expansion. 15. **Trigonometry:** $\tan\theta = \frac{5}{6}$ given. 16. **Vector:** $4\mathbf{i} - 3\mathbf{j} + \mathbf{k}$ as a vector. 17. **Complex number division:** $$\frac{1 + 4i}{3 + i} = \frac{(1 + 4i)(3 - i)}{(3+i)(3 - i)} = \frac{3 - i + 12i - 4i^2}{9 + 1} = \frac{3 + 11i + 4}{10} = \frac{7 + 11i}{10} = 0.7 + 1.1i$$ 18. **Circle equation:** Given $$x^2 + y^2 - 4x + 6y - 12 = 0$$ Complete the square: $$\left(x^2 - 4x + 4\right) + \left(y^2 + 6y + 9\right) = 12 + 4 + 9$$ $$ (x - 2)^2 + (y + 3)^2 = 25$$ A circle with center $(2, -3)$ and radius $5$. --- ### Final answers summary: - Vector cross product $\mathbf{d} \times \mathbf{b} = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$. - $\sin\theta + \cos\theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$ identity verified. - $\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} = \sec \theta - \tan \theta$ assuming $\cos\theta >0$. - Complex division yields $0.7 + 1.1i$. - Circle centered at $(2, -3)$ radius $5$. - Quadratic solutions derived as above.