If the vectors `veca` and `vecb` are perpendicular to each other then a vector `vecv` in terms of `veca` and `vecb` satisfying the equations `vecv.veca=0, vecv.vecb=1` and `[(vecv, veca, vecb)]=1` is

Options

A. `(vecb)/(|vecb|^(2))+(vecaxxvecb)/(|vecaxxvecb|^(2))`
B. `(vecb)/(|vecb|)+(vecaxxvecb)/(|vecaxxvecb|^(2))`
C. `(vecb)/(|vecb|^(2))+(vecaxxvecb)/(|vecaxxvecb|)`
D. none of these

Explanation

To solve the problem, we need to find the vector \(\vec{v}\) in terms of the vectors \(\vec{a}\) and \(\vec{b}\) given the conditions that \(\vec{v} \cdot \vec{a} = 0\), \(\vec{v} \cdot \vec{b} = 1\), and \([\vec{v}, \vec{a}, \vec{b}] = 1\).

### Step 1: Understand the Conditions
Since \(\vec{a}\) and \(\vec{b}\) are perpendicular, we can use the properties of dot products and cross products. The first condition, \(\vec{v} \cdot \vec{a} = 0\), indicates that \(\vec{v}\) is orthogonal to \(\vec{a}\). The second condition, \(\vec{v} \cdot \vec{b} = 1\), indicates that \(\vec{v}\) has a component in the direction of \(\vec{b}\).

### Step 2: Express \(\vec{v}\) in terms of \(\vec{a}\) and \(\vec{b}\)
We can express \(\vec{v}\) as a linear combination of \(\vec{a}\) and \(\vec{b}\) and a term that ensures it is orthogonal to \(\vec{a}\):
\[ \vec{v} = k \vec{b} + m (\vec{a} \times \vec{b}) \]
where \(k\) and \(m\) are scalars to be determined.

### Step 3: Apply the First Condition
Using the first condition \(\vec{v} \cdot \vec{a} = 0\):
\[ (k \vec{b} + m (\vec{a} \times \vec{b})) \cdot \vec{a} = 0 \]
Since \(\vec{a} \cdot \vec{a} = |\vec{a}|^2\) and \(\vec{b} \cdot \vec{a} = 0\) (because they are perpendicular), we have:
\[ m (\vec{a} \times \vec{b}) \cdot \vec{a} = 0 \]
This condition is automatically satisfied because the cross product \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

### Step 4: Apply the Second Condition
Using the second condition \(\vec{v} \cdot \vec{b} = 1\):
\[ (k \vec{b} + m (\vec{a} \times \vec{b})) \cdot \vec{b} = 1 \]
This simplifies to:
\[ k |\vec{b}|^2 = 1 \implies k = \frac{1}{|\vec{b}|^2} \]

### Step 5: Use the Third Condition
The third condition \([\vec{v}, \vec{a}, \vec{b}] = 1\) represents the scalar triple product, which can be expressed as:
\[ [\vec{v}, \vec{a}, \vec{b}] = \vec{v} \cdot (\vec{a} \times \vec{b}) \]
Substituting \(\vec{v} = \frac{1}{|\vec{b}|^2} \vec{b} + m (\vec{a} \times \vec{b})\):
\[ \left(\frac{1}{|\vec{b}|^2} \vec{b} + m (\vec{a} \times \vec{b})\right) \cdot (\vec{a} \times \vec{b}) = 1 \]
The first term evaluates to zero because \(\vec{b} \cdot (\vec{a} \times \vec{b}) = 0\). Thus, we have:
\[ m \cdot |\vec{a} \times \vec{b}| = 1 \implies m = \frac{1}{|\vec{a} \times \vec{b}|} \]

### Final Expression for \(\vec{v}\)
Combining everything, we have:
\[ \vec{v} = \frac{1}{|\vec{b}|^2} \vec{b} + \frac{1}{|\vec{a} \times \vec{b}|} (\vec{a} \times \vec{b}) \]