To express vector $\mathbf{u}$ in trigonometric form, we need to find its magnitude and direction angle.

Step 1: Find the components of vector $\mathbf{u}$

The vector $\mathbf{u}$ has an initial point $P_1(-3, 1)$ and a terminal point $P_2(-10, 7)$. The components of the vector are found by subtracting the coordinates of the initial point from the terminal point:

$\mathbf{u} = (x_2 - x_1, y_2 - y_1) = (-10 - (-3), 7 - 1) = (-7, 6)$

So, $\mathbf{u} = -7\mathbf{i} + 6\mathbf{j}$.

Step 2: Find the magnitude of $\mathbf{u}$

The magnitude $\|\mathbf{u}\|$ is given by:

$\|\mathbf{u}\| = \sqrt{(-7)^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85}$

So, the magnitude $\|\mathbf{u}\|$ is $\sqrt{85}$.

Step 3: Find the direction angle $\theta$

The direction angle $\theta$ of the vector with respect to the positive x-axis is given by:

$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{6}{-7}\right)$

Since the vector is in the second quadrant (where the x-component is negative and the y-component is positive), the angle $\theta$ should be adjusted accordingly. We find the principal value and then adjust for the correct quadrant:

$\theta = \tan^{-1}\left(-\frac{6}{7}\right)$

This gives an angle in the fourth quadrant, so we add $180^\circ$ to place it in the second quadrant:

$\theta = 180^\circ + \tan^{-1}\left(-\frac{6}{7}\right)$

Final Answer:

The vector $\mathbf{u}$ in trigonometric form is:

$\mathbf{u} = \sqrt{85} \left(\cos \theta \mathbf{i} + \sin \theta \mathbf{j}\right)$

where $\theta$ is the angle $180^\circ + \tan^{-1}\left(-\frac{6}{7}\right)$.

Do you need more details on any of these steps or have any other questions?

Here are 5 related questions to consider:

1. How do you convert a vector from trigonometric form back to component form?
2. What are the different ways to express a vector in polar coordinates?
3. How do you find the unit vector in the direction of $\mathbf{u}$?
4. How does the direction angle relate to the components of a vector?
5. What is the significance of the magnitude and direction in physical applications of vectors?

Tip: When working with angles in trigonometric functions, always pay attention to the quadrant in which the vector lies to correctly determine the direction angle.