Let's solve the vector-related problems shown in your image.

Problem 1: Resultant of Two Forces

Given:

• $F_1 = 10 \, \text{N}$ at an angle of 45° from the X-axis
• $F_2 = 8 \, \text{N}$ at an angle of 30° from the X-axis

(a) Magnitude of the Resultant Vector

We need to calculate the resultant of two vectors using vector addition. First, let's break the forces into their X and Y components.

• For $F_1$: $F_{1x} = F_1 \cos(45^\circ), \quad F_{1y} = F_1 \sin(45^\circ)$ $F_{1x} = 10 \cos(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = 7.07 \, \text{N}$ $F_{1y} = 10 \sin(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = 7.07 \, \text{N}$

• For $F_2$: $F_{2x} = F_2 \cos(30^\circ), \quad F_{2y} = F_2 \sin(30^\circ)$ $F_{2x} = 8 \cos(30^\circ) = 8 \times \frac{\sqrt{3}}{2} = 6.93 \, \text{N}$ $F_{2y} = 8 \sin(30^\circ) = 8 \times \frac{1}{2} = 4.00 \, \text{N}$

• Resultant X and Y Components: $R_x = F_{1x} - F_{2x} = 7.07 \, \text{N} - 6.93 \, \text{N} = 0.14 \, \text{N}$ $R_y = F_{1y} + F_{2y} = 7.07 \, \text{N} + 4.00 \, \text{N} = 11.07 \, \text{N}$

• Magnitude of the Resultant Vector: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(0.14)^2 + (11.07)^2} = \sqrt{122.57} = 11.07 \, \text{N}$

(b) Direction of the Resultant Vector

The direction (θ) is given by: $\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{11.07}{0.14}\right) \approx 89.3^\circ \text{ from the positive X-axis}$

(c) Vector Diagram

The resultant vector points almost vertically along the Y-axis due to the small X-component.

Problem 2: Vector Addition and Subtraction

Given:

• $\vec{A} = 3\hat{i} + 4\hat{j}$
• $\vec{B} = 2\hat{i} + \hat{j} - 3\hat{k}$

We perform the following operations:

• (a) $\vec{A} + \vec{B}$: $\vec{A} + \vec{B} = (3 + 2)\hat{i} + (4 + 1)\hat{j} + (0 - 3)\hat{k} = 5\hat{i} + 5\hat{j} - 3\hat{k}$

• (b) $\vec{A} - \vec{B}$: $\vec{A} - \vec{B} = (3 - 2)\hat{i} + (4 - 1)\hat{j} + (0 + 3)\hat{k} = 1\hat{i} + 3\hat{j} + 3\hat{k}$

• (c) $\vec{A} \cdot \vec{B}$: Dot product $\vec{A} \cdot \vec{B} = (3)(2) + (4)(1) + (0)(-3) = 6 + 4 + 0 = 10$

• (d) $\vec{A} \times \vec{B}$: Cross product $\vec{A} \times \vec{B} = \hat{i}((4)(-3) - (0)(1)) - \hat{j}((3)(-3) - (0)(2)) + \hat{k}((3)(1) - (4)(2))$ $= \hat{i}(-12) - \hat{j}(-9) + \hat{k}(-5) = -12\hat{i} + 9\hat{j} - 5\hat{k}$

Problem 3: Dot and Cross Product of Two Vectors

Given:

• $|\vec{A}| = 4$, $|\vec{B}| = 8$, and the angle between them is 30°.

• (a) Dot Product: $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(30^\circ) = 4 \times 8 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.71$

• (b) Cross Product: $|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin(30^\circ) = 4 \times 8 \times \frac{1}{2} = 16$

Would you like to explore any of these calculations further? Here are some related questions for deeper understanding:

1. How do you calculate the cross product using matrix determinants?
2. What are the applications of vector addition in physics?
3. How do you resolve vectors into their components when the angle is not with respect to the X-axis?
4. Can you explain the difference between dot product and cross product geometrically?
5. How does the angle between two vectors affect the magnitude of their cross product?

Tip: Always resolve vectors into their components when dealing with forces acting at angles.