Find the direction angle of vector t = -9i - 10j

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

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

where \( a \) and \( b \) are the components of the vector \( \mathbf{t} = a\mathbf{i} + b\mathbf{j} \).

For the given vector \( \mathbf{t} = -9\mathbf{i} - 10\mathbf{j} \), we have:
- \( a = -9 \)
- \( b = -10 \)

Now, calculate \( \theta \):

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

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

Using a calculator:

\[
\theta \approx \tan^{-1} (1.11) \approx 48.01^\circ
\]

However, since the vector \( \mathbf{t} \) lies in the third quadrant (both \( a \) and \( b \) are negative), we must adjust the angle by adding \( 180^\circ \):

\[
\theta = 48.01^\circ + 180^\circ = 228.01^\circ
\]

Thus, the direction angle of the vector \( \mathbf{t} \) is approximately \( 228.01^\circ \).

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

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Here are 5 related questions for further practice:
1. What is the direction angle of vector \( \mathbf{v} = 5\mathbf{i} - 12\mathbf{j} \)?
2. How would you find the magnitude of vector \( \mathbf{t} = -9\mathbf{i} - 10\mathbf{j} \)?
3. What is the direction angle of a vector that lies in the second quadrant?
4. How do you find the unit vector in the direction of a given vector?
5. How do the signs of the components affect the quadrant of the direction angle?

**Tip:** Always check which quadrant the vector is in after calculating the initial angle, to make necessary adjustments.

Learn how to find the direction angle of the vector t = -9i - 10j using trigonometric functions. The solution includes the inverse tangent formula, vector components, and quadrant adjustment. Suitable for students in Grades 10-12.

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