If `veca` and `vecb` are mutually perpendicular unit vectors, `vecr`is a vector satisfying `vecr.veca =0, vecr.vecb=1` and `(vecr veca vecb)=1`, then `vecr` is:

Options

A. `veca + (veca xx vecb)`
B. `vecb +(veca xx vecb)`
C. `veca + (veca xx vecb)`
D. `veca - vecb+(veca xx vecb)`

Answer

To solve the problem, we need to find the vector \(\vec{r}\) given the conditions involving the unit vectors \(\vec{a}\) and \(\vec{b}\). Let's break down the solution step by step.

### Step 1: Understand the Given Information
We are given:
1. \(\vec{a}\) and \(\vec{b}\) are mutually perpendicular unit vectors.
2. \(\vec{r} \cdot \vec{a} = 0\)
3. \(\vec{r} \cdot \vec{b} = 1\)
4. \((\vec{r} \times \vec{a} \times \vec{b}) = 1\)

### Step 2: Express \(\vec{r}\) in Terms of \(\vec{a}\), \(\vec{b}\), and \(\vec{a} \times \vec{b}\)
Since \(\vec{a}\) and \(\vec{b}\) are mutually perpendicular unit vectors, we can express \(\vec{r}\) as:
\[ \vec{r} = x_1 \vec{a} + x_2 \vec{b} + x_3 (\vec{a} \times \vec{b}) \]
where \(x_1\), \(x_2\), and \(x_3\) are scalars to be determined.

### Step 3: Use the Condition \(\vec{r} \cdot \vec{a} = 0\)
Calculating the dot product:
\[ \vec{r} \cdot \vec{a} = (x_1 \vec{a} + x_2 \vec{b} + x_3 (\vec{a} \times \vec{b})) \cdot \vec{a} \]
This simplifies to:
\[ x_1 (\vec{a} \cdot \vec{a}) + x_2 (\vec{b} \cdot \vec{a}) + x_3 ((\vec{a} \times \vec{b}) \cdot \vec{a}) = 0 \]
Since \(\vec{a} \cdot \vec{a} = 1\), \(\vec{b} \cdot \vec{a} = 0\), and \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\), we have:
\[ x_1 = 0 \]

### Step 4: Use the Condition \(\vec{r} \cdot \vec{b} = 1\)
Now, substituting \(x_1 = 0\):
\[ \vec{r} \cdot \vec{b} = (0 \cdot \vec{a} + x_2 \vec{b} + x_3 (\vec{a} \times \vec{b})) \cdot \vec{b} \]
This simplifies to:
\[ x_2 (\vec{b} \cdot \vec{b}) + x_3 ((\vec{a} \times \vec{b}) \cdot \vec{b}) = 1 \]
Since \(\vec{b} \cdot \vec{b} = 1\) and \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\), we have:
\[ x_2 = 1 \]

### Step 5: Use the Condition \((\vec{r} \times \vec{a} \times \vec{b}) = 1\)
Now we need to evaluate:
\[ \vec{r} \times \vec{a} \times \vec{b} = (0 \cdot \vec{a} + 1 \cdot \vec{b} + x_3 (\vec{a} \times \vec{b})) \times \vec{a} \]
This simplifies to:
\[ \vec{b} \times \vec{a} + x_3 ((\vec{a} \times \vec{b}) \times \vec{a}) \]
Using the vector triple product identity, we have:
\[ (\vec{a} \times \vec{b}) \times \vec{a} = -\vec{b} \]
Thus, we get:
\[ \vec{b} \times \vec{a} - x_3 \vec{b} = 1 \]
Since \(\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}\), we can set the magnitude:
\[ 1 - x_3 = 1 \implies x_3 = 0 \]

### Step 6: Final Expression for \(\vec{r}\)
Substituting \(x_1 = 0\), \(x_2 = 1\), and \(x_3 = 0\) into the expression for \(\vec{r}\):
\[ \vec{r} = 0 \cdot \vec{a} + 1 \cdot \vec{b} + 0 \cdot (\vec{a} \times \vec{b}) = \vec{b} \]

### Conclusion
Thus, the vector \(\vec{r}\) is:
\[ \vec{r} = \vec{b} \]