If `veca, vecb, vecc` are three non-coplanar vectors, then a vector `vecr` satisfying `vecr.veca=vecr.vecb=vecr.vecc=1`, is

Options

A. `vecaxxvecb+vecbxxvecc+veccxxveca`
B. `1/([(veca, vecb, vecc)]){vecaxxvecb+vecbxxvec+veccxxveca}`
C. `[(veca, vecb, vecc)]{vecaxxvecb+vecbxxvecc+vecxxveca}`
D. none of these

Explanation

To find the vector \(\vec{r}\) that satisfies the conditions \(\vec{r} \cdot \vec{a} = \vec{r} \cdot \vec{b} = \vec{r} \cdot \vec{c} = 1\) given that \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors, we can follow these steps:

### Step 1: Understand the implications of non-coplanarity
Since \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar, the vectors \(\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}\) are also non-coplanar. This means we can express any vector \(\vec{r}\) in terms of these cross products.

### Step 2: Express \(\vec{r}\) in terms of cross products
We can express \(\vec{r}\) as:
\[ \vec{r} = X (\vec{a} \times \vec{b}) + Y (\vec{b} \times \vec{c}) + Z (\vec{c} \times \vec{a}) \]
where \(X, Y, Z\) are scalars to be determined.

### Step 3: Take the dot product with \(\vec{a}\)
Taking the dot product of \(\vec{r}\) with \(\vec{a}\):
\[ \vec{r} \cdot \vec{a} = X (\vec{a} \times \vec{b}) \cdot \vec{a} + Y (\vec{b} \times \vec{c}) \cdot \vec{a} + Z (\vec{c} \times \vec{a}) \cdot \vec{a} \]
The terms involving \((\vec{a} \times \vec{b})\) and \((\vec{c} \times \vec{a})\) will be zero because they are perpendicular to \(\vec{a}\). Thus:
\[ \vec{r} \cdot \vec{a} = Y (\vec{b} \times \vec{c}) \cdot \vec{a} \]
Setting this equal to 1 (from the problem statement):
\[ Y (\vec{b} \times \vec{c}) \cdot \vec{a} = 1 \]

### Step 4: Solve for \(Y\)
From the equation above, we can express \(Y\):
\[ Y = \frac{1}{(\vec{b} \times \vec{c}) \cdot \vec{a}} \]

### Step 5: Repeat for \(\vec{b}\) and \(\vec{c}\)
Similarly, taking the dot product of \(\vec{r}\) with \(\vec{b}\):
\[ \vec{r} \cdot \vec{b} = X (\vec{a} \times \vec{b}) \cdot \vec{b} + Y (\vec{b} \times \vec{c}) \cdot \vec{b} + Z (\vec{c} \times \vec{a}) \cdot \vec{b} \]
Again, the terms involving \((\vec{a} \times \vec{b})\) and \((\vec{b} \times \vec{c})\) will be zero, leading to:
\[ \vec{r} \cdot \vec{b} = Z (\vec{c} \times \vec{a}) \cdot \vec{b} = 1 \]
Thus,
\[ Z = \frac{1}{(\vec{c} \times \vec{a}) \cdot \vec{b}} \]

For \(\vec{c}\):
\[ \vec{r} \cdot \vec{c} = X (\vec{a} \times \vec{b}) \cdot \vec{c} + Y (\vec{b} \times \vec{c}) \cdot \vec{c} + Z (\vec{c} \times \vec{a}) \cdot \vec{c} \]
This simplifies to:
\[ \vec{r} \cdot \vec{c} = X (\vec{a} \times \vec{b}) \cdot \vec{c} = 1 \]
Thus,
\[ X = \frac{1}{(\vec{a} \times \vec{b}) \cdot \vec{c}} \]

### Step 6: Substitute back into \(\vec{r}\)
Now we have:
\[ X = \frac{1}{(\vec{a} \times \vec{b}) \cdot \vec{c}}, \quad Y = \frac{1}{(\vec{b} \times \vec{c}) \cdot \vec{a}}, \quad Z = \frac{1}{(\vec{c} \times \vec{a}) \cdot \vec{b}} \]
Substituting these values back into the expression for \(\vec{r}\):
\[ \vec{r} = \frac{1}{(\vec{a} \times \vec{b}) \cdot \vec{c}} (\vec{a} \times \vec{b}) + \frac{1}{(\vec{b} \times \vec{c}) \cdot \vec{a}} (\vec{b} \times \vec{c}) + \frac{1}{(\vec{c} \times \vec{a}) \cdot \vec{b}} (\vec{c} \times \vec{a}) \]

### Final Result
Thus, the vector \(\vec{r}\) is:
\[ \vec{r} = \frac{1}{\text{scalar triple product of } \vec{a}, \vec{b}, \vec{c}} \left( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right) \]