Let's go through the problem step by step:

(i) Calculate the magnitudes $|\mathbf{v}_1|$ and $|\mathbf{v}_2|$:

Given vectors: $\mathbf{v}_1 = 5\hat{i} + 7\hat{j}, \quad \mathbf{v}_2 = 3\hat{i} + 4\hat{j}$

To find the magnitudes, use the formula for the magnitude of a vector: $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$

• For $\mathbf{v}_1 = 5\hat{i} + 7\hat{j}$: $|\mathbf{v}_1| = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74}$

• For $\mathbf{v}_2 = 3\hat{i} + 4\hat{j}$: $|\mathbf{v}_2| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Thus: $|\mathbf{v}_1| = \sqrt{74}, \quad |\mathbf{v}_2| = 5$

(ii) Find the scalar (dot) product, $\mathbf{v}_d = \mathbf{v}_1 \cdot \mathbf{v}_2$, and use it to find the angle between $\mathbf{v}_1$ and $\mathbf{v}_2$:

The dot product formula is: $\mathbf{v}_1 \cdot \mathbf{v}_2 = v_{1x} \cdot v_{2x} + v_{1y} \cdot v_{2y}$

• For $\mathbf{v}_1 = 5\hat{i} + 7\hat{j}$ and $\mathbf{v}_2 = 3\hat{i} + 4\hat{j}$: $\mathbf{v}_1 \cdot \mathbf{v}_2 = (5)(3) + (7)(4) = 15 + 28 = 43$

The angle $\theta$ between $\mathbf{v}_1$ and $\mathbf{v}_2$ can be found using the dot product formula: $\cos(\theta) = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{|\mathbf{v}_1| |\mathbf{v}_2|}$ Substitute values: $\cos(\theta) = \frac{43}{\sqrt{74} \times 5}$ $\cos(\theta) = \frac{43}{5\sqrt{74}} \quad \Rightarrow \quad \theta = \cos^{-1} \left(\frac{43}{5\sqrt{74}}\right)$

(iii) Find the vector (cross) product, $\mathbf{v}_c = \mathbf{v}_1 \times \mathbf{v}_2$, and compute $\mathbf{v}_c \cdot \mathbf{v}_1$ and $\mathbf{v}_c \cdot \mathbf{v}_2$:

For two vectors in 2D, the cross product gives a scalar result in the $k$-direction (z-axis in 3D). The formula for 2D vectors is: $\mathbf{v}_1 \times \mathbf{v}_2 = (v_{1x}v_{2y} - v_{1y}v_{2x})\hat{k}$

• For $\mathbf{v}_1 = 5\hat{i} + 7\hat{j}$ and $\mathbf{v}_2 = 3\hat{i} + 4\hat{j}$: $\mathbf{v}_1 \times \mathbf{v}_2 = (5)(4) - (7)(3) = 20 - 21 = -1\hat{k}$

Thus, $\mathbf{v}_c = -\hat{k}$.

Now compute the dot products: $\mathbf{v}_c \cdot \mathbf{v}_1 = 0, \quad \mathbf{v}_c \cdot \mathbf{v}_2 = 0$ (since $\mathbf{v}_c$ is perpendicular to both $\mathbf{v}_1$ and $\mathbf{v}_2$).

(iv) Evaluate the magnitude of $|\mathbf{v}_1 \times \mathbf{v}_2|$ and use it to find the angle between $\mathbf{v}_1$ and $\mathbf{v}_2$:

The magnitude of the cross product in 2D is simply the absolute value of the scalar result: $|\mathbf{v}_1 \times \mathbf{v}_2| = | -1 | = 1$

Now, the sine of the angle $\theta$ between the vectors is given by: $|\mathbf{v}_1 \times \mathbf{v}_2| = |\mathbf{v}_1| |\mathbf{v}_2| \sin(\theta)$ Substitute values: $1 = \sqrt{74} \times 5 \times \sin(\theta)$ $\sin(\theta) = \frac{1}{5\sqrt{74}}$

Thus, $\theta = \sin^{-1} \left(\frac{1}{5\sqrt{74}}\right)$.

Would you like more details or explanations for any part? Here are some questions to deepen understanding:

1. What is the significance of the scalar (dot) product in vector analysis?
2. How can the angle between two vectors help in physical interpretations?
3. Why is the cross product of two 2D vectors in the $\hat{k}$-direction (z-axis)?
4. Can the dot product of the cross product with the original vectors ever be non-zero in 2D?
5. How would the magnitudes of the vectors affect the angle between them?

Tip: Remember that the cross product in 2D always results in a vector perpendicular to the plane defined by the two original vectors.