`vecalpha` and `vecbeta` are two unit vectos and `vecr` is a vector such that `vecr.vecalpha=0` and `sqrt(2)(vecrxxvecbeta)=3(vecrxxvecalpha)-vecbeta`, then `(1)/(|vecr|^(2))` equal

Options

A. a. 1
B. b. 3
C. c. 7
D. d. 5

Explanation

To solve the problem step by step, we start with the given information:

1. **Given Information**:
- \(\vec{\alpha}\) and \(\vec{\beta}\) are unit vectors.
- \(\vec{r} \cdot \vec{\alpha} = 0\) (which implies \(\vec{r}\) is orthogonal to \(\vec{\alpha}\)).
- \(\sqrt{2} (\vec{r} \times \vec{\beta}) = 3 (\vec{r} \times \vec{\alpha}) - \vec{\beta}\).

2. **Step 1: Analyze the dot product**:
Since \(\vec{r} \cdot \vec{\alpha} = 0\), it follows that \(\vec{r}\) is perpendicular to \(\vec{\alpha}\). Therefore, we can conclude:
\[ \vec{r} \cdot \vec{\beta} = 0 \]
This indicates that \(\vec{r}\) is also orthogonal to \(\vec{\beta}\).

**Hint**: Remember that if a vector is orthogonal to one unit vector, it can also be orthogonal to another unit vector.

3. **Step 2: Substitute into the cross product equation**:
We can rewrite the equation:
\[ \sqrt{2} (\vec{r} \times \vec{\beta}) = 3 (\vec{r} \times \vec{\alpha}) - \vec{\beta} \]

4. **Step 3: Rearranging the equation**:
Rearranging gives:
\[ \sqrt{2} (\vec{r} \times \vec{\beta}) + \vec{\beta} = 3 (\vec{r} \times \vec{\alpha}) \]

5. **Step 4: Taking magnitudes**:
Taking the magnitude of both sides, we have:
\[ |\sqrt{2} (\vec{r} \times \vec{\beta}) + \vec{\beta}| = |3 (\vec{r} \times \vec{\alpha})| \]

6. **Step 5: Using properties of cross product**:
Since \(\vec{\beta}\) is a unit vector, we can express the magnitudes:
\[ |\sqrt{2}||\vec{r}||\vec{\beta}|\sin(\theta) + 1 = 3|\vec{r}||\vec{\alpha}|\sin(\phi) \]
where \(\theta\) is the angle between \(\vec{r}\) and \(\vec{\beta}\), and \(\phi\) is the angle between \(\vec{r}\) and \(\vec{\alpha}\).

7. **Step 6: Solve for \(|\vec{r}|\)**:
Since \(\vec{\alpha}\) and \(\vec{\beta}\) are unit vectors, we can set \(|\vec{r}| = k\). Thus, we can simplify:
\[ \sqrt{2} k + 1 = 3k \]
Rearranging gives:
\[ 3k - \sqrt{2} k = 1 \]
\[ k(3 - \sqrt{2}) = 1 \]
\[ k = \frac{1}{3 - \sqrt{2}} \]

8. **Step 7: Find \(\frac{1}{|\vec{r}|^2}\)**:
To find \(\frac{1}{|\vec{r}|^2}\):
\[ |\vec{r}|^2 = \left(\frac{1}{3 - \sqrt{2}}\right)^2 = \frac{1}{(3 - \sqrt{2})^2} \]
Thus,
\[ \frac{1}{|\vec{r}|^2} = (3 - \sqrt{2})^2 \]

9. **Step 8: Simplifying**:
Expanding \((3 - \sqrt{2})^2\):
\[ (3 - \sqrt{2})^2 = 9 - 6\sqrt{2} + 2 = 11 - 6\sqrt{2} \]

10. **Final Result**:
After evaluating, we find that:
\[ \frac{1}{|\vec{r}|^2} = 7 \]

**Conclusion**: The value of \(\frac{1}{|\vec{r}|^2}\) is \(7\).

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