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

Explanation

To solve the problem, we need to find a vector \(\vec{v}\) in terms of vectors \(\vec{a}\) and \(\vec{b}\) that satisfies the following conditions:

1. \(\vec{v} \cdot \vec{a} = 0\)
2. \(\vec{v} \cdot \vec{b} = 1\)
3. \([\vec{v}, \vec{a}, \vec{b}] = 1\)

Given that \(\vec{a}\) and \(\vec{b}\) are perpendicular, we can proceed as follows:

### Step 1: Understand the implications of the conditions

Since \(\vec{a}\) and \(\vec{b}\) are perpendicular, we know that:

\[ \vec{a} \cdot \vec{b} = 0 \]

This means that the dot product of \(\vec{v}\) with \(\vec{a}\) being zero indicates that \(\vec{v}\) is perpendicular to \(\vec{a}\).

### Step 2: Express \(\vec{v}\) in terms of \(\vec{a}\) and \(\vec{b}\)

Since \(\vec{v}\) is perpendicular to \(\vec{a}\), we can express \(\vec{v}\) as a linear combination of \(\vec{b}\) and a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\). We can write:

\[ \vec{v} = x \vec{b} + y (\vec{a} \times \vec{b}) \]

where \(x\) and \(y\) are scalars.

### Step 3: Apply the second condition \(\vec{v} \cdot \vec{b} = 1\)

Now, we substitute \(\vec{v}\) into the second condition:

\[ \vec{v} \cdot \vec{b} = (x \vec{b} + y (\vec{a} \times \vec{b})) \cdot \vec{b} = x (\vec{b} \cdot \vec{b}) + y ((\vec{a} \times \vec{b}) \cdot \vec{b}) \]

Since \((\vec{a} \times \vec{b})\) is perpendicular to \(\vec{b}\), we have:

\[ (\vec{a} \times \vec{b}) \cdot \vec{b} = 0 \]

Thus, the equation simplifies to:

\[ x |\vec{b}|^2 = 1 \implies x = \frac{1}{|\vec{b}|^2} \]

### Step 4: Substitute \(x\) back into \(\vec{v}\)

Now we can substitute \(x\) back into the expression for \(\vec{v}\):

\[ \vec{v} = \frac{1}{|\vec{b}|^2} \vec{b} + y (\vec{a} \times \vec{b}) \]

### Step 5: Apply the third condition \([\vec{v}, \vec{a}, \vec{b}] = 1\)

The scalar triple product \([\vec{v}, \vec{a}, \vec{b}]\) can be computed as:

\[ [\vec{v}, \vec{a}, \vec{b}] = \left(\frac{1}{|\vec{b}|^2} \vec{b} + y (\vec{a} \times \vec{b}), \vec{a}, \vec{b}\right) \]

Calculating this gives:

\[ [\vec{v}, \vec{a}, \vec{b}] = \frac{1}{|\vec{b}|^2} [\vec{b}, \vec{a}, \vec{b}] + y [(\vec{a} \times \vec{b}), \vec{a}, \vec{b}] \]

The first term is zero because \([\vec{b}, \vec{a}, \vec{b}] = 0\). The second term simplifies to:

\[ y |\vec{a}| |\vec{b}| \sin(\theta) = y |\vec{a}| |\vec{b}| \]

where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). Since \(\vec{a}\) and \(\vec{b}\) are perpendicular, \(\sin(\theta) = 1\).

Setting this equal to 1 gives:

\[ y |\vec{a}| |\vec{b}| = 1 \implies y = \frac{1}{|\vec{a}| |\vec{b}|} \]

### Final Expression for \(\vec{v}\)

Substituting \(y\) back into the expression for \(\vec{v}\):

\[ \vec{v} = \frac{1}{|\vec{b}|^2} \vec{b} + \frac{1}{|\vec{a}| |\vec{b}|} (\vec{a} \times \vec{b}) \]

This is the required vector \(\vec{v}\).