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To determine the value of \( k \) so that vectors \( \vec{a} = \langle -2, 3, -3 \rangle \) and \( \vec{b} = \langle -5, 3, k \rangle \) are orthogonal, we use the concept of the **dot product**. Two vectors are orthogonal if their dot product is zero.

1. **Calculate the dot product**:
   \[
   \vec{a} \cdot \vec{b} = (-2)(-5) + (3)(3) + (-3)(k)
   \]

2. **Expand and simplify**:
   \[
   \vec{a} \cdot \vec{b} = 10 + 9 - 3k
   \]

3. **Set the dot product equal to zero for orthogonality**:
   \[
   10 + 9 - 3k = 0
   \]

4. **Solve for \( k \)**:
   \[
   19 - 3k = 0
   \]
   \[
   -3k = -19
   \]
   \[
   k = \frac{19}{3}
   \]

So, the value of \( k \) that makes \( \vec{a} \) and \( \vec{b} \) orthogonal is:
\[
k = \frac{19}{3}
\]

Would you like further details on the solution or have any questions?

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Here are 5 related questions that might expand your understanding:

1. What is the dot product of two vectors, and how is it used in geometry?
2. Why does a zero dot product imply that two vectors are orthogonal?
3. How would you find the angle between two non-orthogonal vectors using their dot product?
4. Can two vectors be parallel if their dot product is zero?
5. What other conditions (besides orthogonality) can be determined using the dot product?

**Tip:** When solving vector problems, always check whether simplifying the components (like using symmetry or common factors) might reduce calculations.

Let \( \vec{a} = \langle -2, 3, -3 \rangle \) and \( \vec{b} = \langle -5, 3, k \rangle \). Find \( k \) so that \( \vec{a} \) and \( \vec{b} \) will be orthogonal.

Solve for the value of \( k \) that makes vectors \( \vec{a} = \langle -2, 3, -3 \rangle \) and \( \vec{b} = \langle -5, 3, k \rangle \) orthogonal. This problem involves key concepts in vector algebra, including the dot product and the condition for vector orthogonality, suitable for students in Grades 10-12.

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