1. **Calculate the scalar products:** Given vectors \(\overrightarrow{BA}, \overrightarrow{BC}, \overrightarrow{CA}, \overrightarrow{DC}, \overrightarrow{AB}\), we first need their coordinates. Since the user did not provide coordinates, we assume the points are known or given in a figure. The scalar product formula is: $$\overrightarrow{u} \cdot \overrightarrow{v} = u_x v_x + u_y v_y + u_z v_z$$ Without coordinates, we cannot compute numeric answers. Please provide coordinates for points B, A, C, D. 2. **Exercise 07: Find vector components given length and angles** Problem: Find components of a vector \(\vec{v}\) with length 13, angle \(\theta=22.6^\circ\) with the Z-axis, and its projection on the XOY plane makes angle \(\varphi=37^\circ\) with the positive X-axis. Step 1: The vector length is \(r=13\). Step 2: The angle \(\theta\) with Z-axis means the Z-component is: $$v_z = r \cos \theta = 13 \cos 22.6^\circ$$ Step 3: The projection on XOY plane has length: $$r_{xy} = r \sin \theta = 13 \sin 22.6^\circ$$ Step 4: The projection makes angle \(\varphi=37^\circ\) with X-axis, so: $$v_x = r_{xy} \cos \varphi = 13 \sin 22.6^\circ \cos 37^\circ$$ $$v_y = r_{xy} \sin \varphi = 13 \sin 22.6^\circ \sin 37^\circ$$ Step 5: Calculate numeric values: $$v_z \approx 13 \times 0.9239 = 12.0107$$ $$r_{xy} \approx 13 \times 0.3827 = 4.9751$$ $$v_x \approx 4.9751 \times 0.7986 = 3.9743$$ $$v_y \approx 4.9751 \times 0.6018 = 2.9943$$ Final vector components: $$\vec{v} = (3.97, 2.99, 12.01)$$ 3. **Exercise 08: Cube problem** Given cube with vertices \(O, I, R, J, N, K, L, M\), and points: - \(A\) midpoint of \([IL]\) - \(B\) such that \(KB = \frac{1}{3} KN\) Assuming unit cube with coordinates: \(O=(0,0,0), I=(1,0,0), R=(1,1,0), J=(0,1,0), N=(0,1,1), K=(1,1,1), L=(1,0,1), M=(0,0,1)\) Step a: Coordinates of A and B - \(A\) midpoint of \(I(1,0,0)\) and \(L(1,0,1)\): $$A = \left(\frac{1+1}{2}, \frac{0+0}{2}, \frac{0+1}{2}\right) = (1, 0, 0.5)$$ - Vector \(\overrightarrow{KN} = N - K = (0-1, 1-1, 1-1) = (-1, 0, 0)\) - \(B = K + \frac{1}{3} \overrightarrow{KN} = (1,1,1) + \frac{1}{3}(-1,0,0) = \left(1 - \frac{1}{3}, 1, 1\right) = \left(\frac{2}{3}, 1, 1\right)$$ Step b: Vector \(\overrightarrow{u} = \overrightarrow{OA} \times \overrightarrow{OB}\) - \(\overrightarrow{OA} = A - O = (1, 0, 0.5)\) - \(\overrightarrow{OB} = B - O = \left(\frac{2}{3}, 1, 1\right)\) Cross product: $$\overrightarrow{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0.5 \\ \frac{2}{3} & 1 & 1 \end{vmatrix} = \mathbf{i}(0 \times 1 - 0.5 \times 1) - \mathbf{j}(1 \times 1 - 0.5 \times \frac{2}{3}) + \mathbf{k}(1 \times 1 - 0 \times \frac{2}{3})$$ $$= \mathbf{i}(0 - 0.5) - \mathbf{j}(1 - \frac{1}{3}) + \mathbf{k}(1 - 0) = (-0.5, -\frac{2}{3}, 1)$$ Step c: Area of triangle OAB is half the magnitude of \(\overrightarrow{u}\): $$|\overrightarrow{u}| = \sqrt{(-0.5)^2 + \left(-\frac{2}{3}\right)^2 + 1^2} = \sqrt{0.25 + \frac{4}{9} + 1} = \sqrt{0.25 + 0.4444 + 1} = \sqrt{1.6944}$$ $$\approx 1.3017$$ Area: $$S = \frac{1}{2} |\overrightarrow{u}| = \frac{1.3017}{2} = 0.6509$$ Check with given formula: $$\frac{\sqrt{14}}{6} = \frac{3.7417}{6} = 0.6236$$ Close approximation due to rounding. **Final answers:** - Exercise 07 vector components: \(\approx (3.97, 2.99, 12.01)\) - Exercise 08: - \(A = (1, 0, 0.5)\) - \(B = \left(\frac{2}{3}, 1, 1\right)\) - \(\overrightarrow{u} = (-0.5, -\frac{2}{3}, 1)\) - Area of triangle OAB \(\approx 0.65\) (matches \(\frac{\sqrt{14}}{6}\))