1. **Problem 27:** Find the magnitude of the resultant vector from three vectors along the x, y, and z axes with magnitudes 25, 35, and 40 respectively. 2. The resultant vector $ \vec{R} $ is the vector sum: $$\vec{R} = 25\hat{i} + 35\hat{j} + 40\hat{k}$$ 3. The magnitude of $ \vec{R} $ is given by: $$|\vec{R}| = \sqrt{25^2 + 35^2 + 40^2}$$ 4. Calculate each square: $$25^2 = 625, \quad 35^2 = 1225, \quad 40^2 = 1600$$ 5. Sum the squares: $$625 + 1225 + 1600 = 3450$$ 6. Take the square root: $$|\vec{R}| = \sqrt{3450} \approx 58.7$$ --- 1. **Problem 28:** Find the tension $ T $ in a cable holding a 1000 kg mass at an angle of 35° from the horizontal beam. 2. The weight force $ W $ is: $$W = mg = 1000 \times 9.8 = 9800$$ 3. The tension $ T $ in the cable supports the weight vertically, so: $$T \sin 35^\circ = W$$ 4. Solve for $ T $: $$T = \frac{W}{\sin 35^\circ}$$ 5. Calculate $ \sin 35^\circ \approx 0.574 $: $$T = \frac{9800}{0.574} \approx 17085.8$$ --- 1. **Problem 26:** Determine which door diagram yields the greatest torque around the hinges. 2. Torque $ \tau = rF \sin \theta $, where $ r $ is the distance from the hinge, $ F $ is force magnitude, and $ \theta $ is the angle between force and lever arm. 3. Diagram A applies full force at outer half with an angle upward-right. 4. Diagram B applies full force straight upward near outer half. 5. Diagram C applies full force angled downward-right at far end. 6. Diagram D applies half force at far end angled upward-right. 7. Since torque depends on force magnitude and lever arm, Diagram C or A likely yield greatest torque due to full force and maximum lever arm. 8. Without exact values, Diagram C (force at far end angled downward-right) yields greatest torque. --- 1. **Problem 25:** Express vectors $ \vec{CD} $ and $ \vec{CE} $ in terms of $ \vec{u} = \vec{AB} $ and $ \vec{v} = \vec{AE} $, given $ BD \parallel AE $ and $ AB = 5 BC $. 2. Since $ BD \parallel AE $, $ \vec{BD} = k \vec{v} $ for some scalar $ k $. 3. Using vector addition and given relations, express $ \vec{CD} = \vec{BD} - \vec{BC} $. 4. Since $ AB = 5 BC $, $ \vec{BC} = \frac{1}{5} \vec{u} $. 5. Therefore: $$\vec{CD} = k \vec{v} - \frac{1}{5} \vec{u}$$ 6. For $ \vec{CE} $, use: $$\vec{CE} = \vec{CA} + \vec{AE} = -\vec{u} + \vec{v}$$ --- 1. **Problem 23:** Find the vector equation of the plane through points $ (2,0,0), (0,4,0), (0,0,5) $. 2. Two vectors on the plane are: $$\vec{AB} = (0-2,4-0,0-0) = (-2,4,0)$$ $$\vec{AC} = (0-2,0-0,5-0) = (-2,0,5)$$ 3. Parametric form: $$[x,y,z] = [2,0,0] + s(-2,4,0) + t(-2,0,5)$$ 4. This matches the first option with rearranged parameters: $$[x,y,z] = [2,0,0] + t[2,-4,0] + s[0,-2,5]$$ 5. So the correct vector equation is: $$[x,y,z] = [2,0,0] + t[2,-4,0] + s[0,-2,5]$$ --- 1. **Problem 24:** Find the distance between two skew lines: $$l_1: \vec{r} = (5,-4,-2) + s(1,2,3)$$ $$l_2: \vec{r} = (2,0,1) + t(2,-1,1)$$ 2. The distance $ d $ between skew lines is: $$d = \frac{|(\vec{P_2} - \vec{P_1}) \cdot (\vec{v_1} \times \vec{v_2})|}{|\vec{v_1} \times \vec{v_2}|}$$ 3. Compute $ \vec{P_2} - \vec{P_1} = (2-5, 0+4, 1+2) = (-3,4,3) $$ 4. Compute cross product: $$\vec{v_1} = (1,2,3), \quad \vec{v_2} = (2,-1,1)$$ $$\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{vmatrix} = (2 \times 1 - 3 \times (-1))\hat{i} - (1 \times 1 - 3 \times 2)\hat{j} + (1 \times (-1) - 2 \times 2)\hat{k}$$ $$= (2 + 3)\hat{i} - (1 - 6)\hat{j} + (-1 - 4)\hat{k} = 5\hat{i} + 5\hat{j} - 5\hat{k}$$ 5. Compute numerator: $$|(-3,4,3) \cdot (5,5,-5)| = |-15 + 20 - 15| = |-10| = 10$$ 6. Compute denominator: $$|\vec{v_1} \times \vec{v_2}| = \sqrt{5^2 + 5^2 + (-5)^2} = \sqrt{75} = 5\sqrt{3} \approx 8.66$$ 7. Distance: $$d = \frac{10}{8.66} \approx 1.15$$ --- **Final answers:** - Problem 27: 58.7 - Problem 28: 17085.8 - Problem 26: Diagram C - Problem 25: \( \vec{CD} = k \vec{v} - \frac{1}{5} \vec{u} $, $ \vec{CE} = -\vec{u} + \vec{v} $ - Problem 23: $$[x,y,z] = [2,0,0] + t[2,-4,0] + s[0,-2,5]$$ - Problem 24: 1.15