If the magnitude of the moment about the pont `hatj+hatk` of a force `hati+alphahatj-hatk` acting through the point `hati+hatj` is`sqrt(8)`, then the value of `alpha` is

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To solve the problem, we need to find the value of \( \alpha \) given that the magnitude of the moment about the point \( \hat{j} + \hat{k} \) of a force \( \hat{i} + \alpha \hat{j} - \hat{k} \) acting through the point \( \hat{i} + \hat{j} \) is \( \sqrt{8} \).

### Step-by-Step Solution:

1. **Identify the Points and Vectors:**
- Let point A be \( \hat{i} + \hat{j} \).
- Let point B be \( \hat{j} + \hat{k} \).
- The force vector \( \mathbf{F} \) is given as \( \hat{i} + \alpha \hat{j} - \hat{k} \).

2. **Calculate the Position Vector from A to B:**
- The vector \( \overrightarrow{BA} \) can be calculated as:
\[ \overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = (\hat{i} + \hat{j}) - (\hat{j} + \hat{k}) = \hat{i} + \hat{j} - \hat{j} - \hat{k} = \hat{i} - \hat{k} \]
- Thus, \( \overrightarrow{BA} = \hat{i} - \hat{k} \).

3. **Set Up the Cross Product:**
- The moment \( \mathbf{M} \) about point B is given by:
\[ \mathbf{M} = \overrightarrow{BA} \times \mathbf{F} \]
- We can express this as:
\[ \mathbf{M} = (\hat{i} - \hat{k}) \times (\hat{i} + \alpha \hat{j} - \hat{k}) \]

4. **Calculate the Cross Product:**
- Using the determinant form for the cross product:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 1 & \alpha & -1 \end{vmatrix} \]
- Expanding this determinant:
\[ \mathbf{M} = \hat{i} \begin{vmatrix} 0 & -1 \\ \alpha & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 0 \\ 1 & \alpha \end{vmatrix} \]
- This results in:
\[ \mathbf{M} = \hat{i} (0 - (-\alpha)) - \hat{j} (1 - (-1)) + \hat{k} (\alpha - 0) \]
- Simplifying gives:
\[ \mathbf{M} = \alpha \hat{i} - 2 \hat{j} + \alpha \hat{k} \]

5. **Calculate the Magnitude of the Moment:**
- The magnitude of the moment vector \( \mathbf{M} \) is given by:
\[ |\mathbf{M}| = \sqrt{\alpha^2 + (-2)^2 + \alpha^2} = \sqrt{2\alpha^2 + 4} \]

6. **Set the Magnitude Equal to Given Value:**
- We know from the problem that this magnitude equals \( \sqrt{8} \):
\[ \sqrt{2\alpha^2 + 4} = \sqrt{8} \]
- Squaring both sides gives:
\[ 2\alpha^2 + 4 = 8 \]
- Rearranging yields:
\[ 2\alpha^2 = 4 \implies \alpha^2 = 2 \implies \alpha = \sqrt{2} \text{ or } \alpha = -\sqrt{2} \]

7. **Final Answer:**
- The value of \( \alpha \) is \( \sqrt{2} \) or \( -\sqrt{2} \).