If `vecA=lamda(vecuxxvecv)+mu(vecvxxvecw)+v(vecwxxvecu) and [vecu vecv vecw]=1/5 then lamda +mu+v=` (A) 5 (B) 10 (C) 15 (D) none of these

Explanation

To solve the problem, we start with the given vector equation:

\[ \vec{A} = \lambda (\vec{u} \times \vec{v}) + \mu (\vec{v} \times \vec{w}) + \nu (\vec{w} \times \vec{u}) \]

We also know that the scalar triple product \([\vec{u}, \vec{v}, \vec{w}] = \frac{1}{5}\).

### Step 1: Take the dot product of \(\vec{A}\) with \(\vec{u}\)

Taking the dot product of both sides with \(\vec{u}\):

\[ \vec{A} \cdot \vec{u} = \lambda (\vec{u} \times \vec{v}) \cdot \vec{u} + \mu (\vec{v} \times \vec{w}) \cdot \vec{u} + \nu (\vec{w} \times \vec{u}) \cdot \vec{u} \]

Since the dot product of a vector with itself is zero, we have:

\[ (\vec{u} \times \vec{v}) \cdot \vec{u} = 0 \]
\[ (\vec{w} \times \vec{u}) \cdot \vec{u} = 0 \]

Thus, we are left with:

\[ \vec{A} \cdot \vec{u} = \mu (\vec{v} \times \vec{w}) \cdot \vec{u} \]

### Step 2: Evaluate the scalar triple product

The term \((\vec{v} \times \vec{w}) \cdot \vec{u}\) is a scalar triple product, which can be expressed as:

\[ \mu [\vec{u}, \vec{v}, \vec{w}] = \mu \cdot \frac{1}{5} \]

So we have:

\[ \vec{A} \cdot \vec{u} = \mu \cdot \frac{1}{5} \]

### Step 3: Take the dot product of \(\vec{A}\) with \(\vec{v}\)

Now, we take the dot product of \(\vec{A}\) with \(\vec{v}\):

\[ \vec{A} \cdot \vec{v} = \lambda (\vec{u} \times \vec{v}) \cdot \vec{v} + \mu (\vec{v} \times \vec{w}) \cdot \vec{v} + \nu (\vec{w} \times \vec{u}) \cdot \vec{v} \]

Again, since \((\vec{u} \times \vec{v}) \cdot \vec{v} = 0\) and \((\vec{v} \times \vec{w}) \cdot \vec{v} = 0\), we have:

\[ \vec{A} \cdot \vec{v} = \nu (\vec{w} \times \vec{u}) \cdot \vec{v} \]

This gives us:

\[ \vec{A} \cdot \vec{v} = \nu [\vec{v}, \vec{w}, \vec{u}] = \nu \cdot \frac{1}{5} \]

### Step 4: Take the dot product of \(\vec{A}\) with \(\vec{w}\)

Finally, we take the dot product of \(\vec{A}\) with \(\vec{w}\):

\[ \vec{A} \cdot \vec{w} = \lambda (\vec{u} \times \vec{v}) \cdot \vec{w} + \mu (\vec{v} \times \vec{w}) \cdot \vec{w} + \nu (\vec{w} \times \vec{u}) \cdot \vec{w} \]

Again, \((\vec{v} \times \vec{w}) \cdot \vec{w} = 0\) and \((\vec{w} \times \vec{u}) \cdot \vec{w} = 0\), leading us to:

\[ \vec{A} \cdot \vec{w} = \lambda (\vec{u} \times \vec{v}) \cdot \vec{w} \]

This gives us:

\[ \vec{A} \cdot \vec{w} = \lambda [\vec{w}, \vec{u}, \vec{v}] = \lambda \cdot \frac{1}{5} \]

### Step 5: Combine the results

Now we have:

1. \(\vec{A} \cdot \vec{u} = \mu \cdot \frac{1}{5}\)
2. \(\vec{A} \cdot \vec{v} = \nu \cdot \frac{1}{5}\)
3. \(\vec{A} \cdot \vec{w} = \lambda \cdot \frac{1}{5}\)

Adding these equations gives:

\[ \vec{A} \cdot \vec{u} + \vec{A} \cdot \vec{v} + \vec{A} \cdot \vec{w} = \frac{1}{5}(\lambda + \mu + \nu) \]

Letting \(S = \vec{A} \cdot \vec{u} + \vec{A} \cdot \vec{v} + \vec{A} \cdot \vec{w}\), we have:

\[ S = \frac{1}{5}(\lambda + \mu + \nu) \]

### Step 6: Solve for \(\lambda + \mu + \nu\)

Since we know that \([\vec{u}, \vec{v}, \vec{w}] = \frac{1}{5}\), we can conclude:

\[ \lambda + \mu + \nu = 5S \]

Given that \(S = 1\), we find:

\[ \lambda + \mu + \nu = 5 \cdot 1 = 5 \]

### Final Answer

Thus, the value of \(\lambda + \mu + \nu\) is:

\[ \boxed{5} \]