https://file.aiquickdraw.com/mathbot/file/173049854268989z4cw5e.jpg

To solve this problem, let's go through each step carefully.

1. **Identify the Given Vector and Points:**
   - The force vector is given by \( \vec{F} = L \hat{i} - 2 \hat{j} \).
   - Point \( A \) (the point where the force acts) is \( (5, 2) \).
   - Point \( B \) (about which the moment is calculated) is \( (7, -4) \).
   - The moment vector \( \vec{M}_B \) about point \( B \) is given as \( 20 \hat{k} \).

2. **Moment Calculation Formula:**
   The moment of a force about a point \( B \) is given by:
   \[
   \vec{M}_B = \vec{r}_{AB} \times \vec{F}
   \]
   where \( \vec{r}_{AB} \) is the position vector from point \( B \) to point \( A \), and \( \vec{F} \) is the force vector.

3. **Find \( \vec{r}_{AB} \):**
   - The position vector \( \vec{r}_{AB} \) from \( B(7, -4) \) to \( A(5, 2) \) is:
     \[
     \vec{r}_{AB} = (5 - 7) \hat{i} + (2 - (-4)) \hat{j} = -2 \hat{i} + 6 \hat{j}
     \]

4. **Calculate the Moment \( \vec{M}_B = \vec{r}_{AB} \times \vec{F} \):**
   The cross product in two dimensions (resulting in a moment along \( \hat{k} \)) can be computed as:
   \[
   \vec{M}_B = (x_1 y_2 - y_1 x_2) \hat{k}
   \]
   where \( \vec{r}_{AB} = x_1 \hat{i} + y_1 \hat{j} \) and \( \vec{F} = x_2 \hat{i} + y_2 \hat{j} \).

   Substitute:
   - \( \vec{r}_{AB} = -2 \hat{i} + 6 \hat{j} \)
   - \( \vec{F} = L \hat{i} - 2 \hat{j} \)

   Then,
   \[
   \vec{M}_B = (-2)(-2) - (6)(L) = 4 - 6L
   \]

5. **Equate to the Given Moment:**
   Since we know \( \vec{M}_B = 20 \hat{k} \), we can set up the equation:
   \[
   4 - 6L = 20
   \]

6. **Solve for \( L \):**
   \[
   -6L = 20 - 4
   \]
   \[
   -6L = 16
   \]
   \[
   L = -\frac{16}{6} = -\frac{8}{3}
   \]

7. **Conclusion:**
   The correct answer is:
   \[
   \boxed{-\frac{8}{3}}
   \]
   Therefore, the answer is (c) \( -\frac{8}{3} \).

Would you like more details on any of the steps, or have any questions? 

Here are some related questions that expand on this concept:
1. How is the direction of the cross product determined in three dimensions?
2. What is the physical significance of a moment vector in mechanics?
3. How would the calculation change if \( \vec{F} \) acted at a different point than \( A \)?
4. What if the moment vector was given in terms of other directions (e.g., \( \hat{i} \) or \( \hat{j} \))?
5. How can we visualize a cross product in two dimensions?

**Tip:** Always keep track of the direction of vectors, especially when computing moments, as incorrect directions can lead to sign errors.

If F = L î - 2 ĵ acts at the point A(5, 2) and the moment vector of F about the point B(7, -4) equals 20 k̂, then the value of L is...

Learn how to find the value of L when a force vector F acts at point A and the moment vector about point B is given. This problem involves vector concepts, cross products, and moment calculations, suitable for undergraduate students in engineering or physics.

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