Let `oversetto a=3hati+2hatj+xhatk` and `oversettob=hati-hatj+hatk`
Some real x. Then `|oversettoaxxoversettob|=r` is possible if:

Options

A. `sqrt(3/2)lt r le3sqrt(3/2)`
B. `rge5sqrt(3/2)`
C. `0ltrlesqrt(3/2)`
D. `0ltrlesqrt(3/2)`

Answer

To solve the problem, we need to find the conditions under which the magnitude of the cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is equal to \( r \).

Given:
\[ \mathbf{A} = 3\hat{i} + 2\hat{j} + x\hat{k} \]
\[ \mathbf{B} = \hat{i} - \hat{j} + \hat{k} \]

### Step 1: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)

The cross product can be calculated using the determinant of the following matrix:

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & -1 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ = \hat{i} \begin{vmatrix} 2 & x \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & x \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ 2 \cdot 1 - (-1) \cdot x = 2 + x \]

2. For \( \hat{j} \):
\[ 3 \cdot 1 - 1 \cdot x = 3 - x \]

3. For \( \hat{k} \):
\[ 3 \cdot (-1) - 2 \cdot 1 = -3 - 2 = -5 \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = (2 + x)\hat{i} - (3 - x)\hat{j} - 5\hat{k} \]

### Step 2: Find the magnitude of \( \mathbf{A} \times \mathbf{B} \)

The magnitude is given by:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(2 + x)^2 + (3 - x)^2 + (-5)^2} \]

Calculating this:
\[ = \sqrt{(2 + x)^2 + (3 - x)^2 + 25} \]

Expanding the squares:
\[ = \sqrt{(4 + 4x + x^2) + (9 - 6x + x^2) + 25} \]
\[ = \sqrt{2x^2 - 2x + 38} \]

### Step 3: Set the magnitude equal to \( r \)

We set:
\[ \sqrt{2x^2 - 2x + 38} = r \]

Squaring both sides:
\[ 2x^2 - 2x + 38 = r^2 \]

### Step 4: Analyze the quadratic equation

This is a quadratic equation in the form:
\[ 2x^2 - 2x + (38 - r^2) = 0 \]

For real solutions of \( x \), the discriminant must be non-negative:
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot (38 - r^2) \geq 0 \]
\[ 4 - 8(38 - r^2) \geq 0 \]
\[ 4 - 304 + 8r^2 \geq 0 \]
\[ 8r^2 \geq 300 \]
\[ r^2 \geq 37.5 \]
\[ r \geq \frac{5\sqrt{3}}{2} \]

### Conclusion

Thus, the condition for \( | \mathbf{A} \times \mathbf{B} | = r \) to be possible is:
\[ r \geq \frac{5\sqrt{3}}{2} \]