To find the direction angle \(\theta\) of the vector \(\mathbf{t} = 9i - 10j\), we can use the following formula:

\[
\theta = \tan^{-1}\left(\frac{y}{x}\right)
\]

Where:
- \(x = 9\) (the coefficient of \(i\))
- \(y = -10\) (the coefficient of \(j\))

\[
\theta = \tan^{-1}\left(\frac{-10}{9}\right)
\]

Now, calculate:

\[
\theta = \tan^{-1}\left(\frac{-10}{9}\right) \approx \tan^{-1}(-1.111) \approx -48.37^\circ
\]

Since the vector is in the fourth quadrant (positive \(x\), negative \(y\)), we need to adjust the angle to its positive equivalent:

\[
\theta = 360^\circ - 48.37^\circ = 311.63^\circ
\]

So, the direction angle of \(\mathbf{t} = 9i - 10j\) is approximately \(311.63^\circ\).

Would you like more details on this calculation or have any questions?

Here are 5 related questions:
1. What is the magnitude of vector \(\mathbf{t} = 9i - 10j\)?
2. How do you find the direction angle of a 3D vector?
3. Can you derive the direction angles for a vector in each quadrant?
4. What are direction cosines, and how are they related to direction angles?
5. How do you convert between polar and rectangular coordinates?

**Tip:** The direction angle is always measured counterclockwise from the positive x-axis, unless otherwise specified.

Discover how to calculate the direction angle of the vector t = 9i - 10j using trigonometric principles. This step-by-step solution explains how to apply the formula θ = tan⁻¹(y/x), adjust for the correct quadrant, and determine that the direction angle is approximately 311.63°. Suitable for high school students in Grades 10-12.

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