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To find a unit vector in the direction of the vector \(\mathbf{v} = -12\mathbf{i} + 15\mathbf{j} - 16\mathbf{k}\), we need to follow these steps:

1. **Calculate the Magnitude of \(\mathbf{v}\):**
   The magnitude of \(\mathbf{v}\), denoted as \(|\mathbf{v}|\), is given by:
   \[
   |\mathbf{v}| = \sqrt{(-12)^2 + 15^2 + (-16)^2}
   \]
   \[
   |\mathbf{v}| = \sqrt{144 + 225 + 256} = \sqrt{625} = 25
   \]

2. **Divide Each Component by the Magnitude:**
   To find the unit vector in the direction of \(\mathbf{v}\), we divide each component of \(\mathbf{v}\) by \( |\mathbf{v}| \):
   \[
   \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{-12}{25} \mathbf{i} + \frac{15}{25} \mathbf{j} + \frac{-16}{25} \mathbf{k}
   \]
   
3. **Simplify Each Component:**
   This gives us:
   \[
   \mathbf{u} = -\frac{12}{25} \mathbf{i} + \frac{3}{5} \mathbf{j} - \frac{16}{25} \mathbf{k}
   \]

Therefore, the unit vector in the direction of \(\mathbf{v}\) is:
\[
\mathbf{u} = -\frac{12}{25} \mathbf{i} + \frac{3}{5} \mathbf{j} - \frac{16}{25} \mathbf{k}
\]

Would you like further details or have any questions?

Here are five related questions to expand on this topic:
1. How do you determine the direction of a vector given two points?
2. What is the significance of a unit vector in physics or engineering?
3. How would you normalize a vector if it were in 4-dimensional space?
4. Can a zero vector have a unit vector in its direction?
5. How would this process differ if you needed a unit vector perpendicular to \(\mathbf{v}\)?

**Tip:** When finding the direction of a vector, ensure to first confirm its magnitude to normalize it accurately.

Find a unit vector in the direction of \( \mathbf{v} = -12\mathbf{i} + 15\mathbf{j} - 16\mathbf{k} \).

Learn how to find a unit vector in the direction of a vector \( \mathbf{v} = -12\mathbf{i} + 15\mathbf{j} - 16\mathbf{k} \) by calculating its magnitude and dividing each component by it. This problem is suitable for students in Grades 10-12.

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