If `vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` and `[veca vecb vecc]=1/8`, then `x+y+z=`

Options

A. `8vec(alpha).(veca+vecb+vecc)`
B. `vec(alpha).(veca+vecb+vecc)`
C. `8(veca+vecb+vecc)`
D. None of these

Explanation

To solve the problem, we start with the given expression for the vector \(\vec{\alpha}\):

\[ \vec{\alpha} = x(\vec{a} \times \vec{b}) + y(\vec{b} \times \vec{c}) + z(\vec{c} \times \vec{a}) \]

We also have the condition that:

\[ [\vec{a}, \vec{b}, \vec{c}] = \frac{1}{8} \]

This notation represents the scalar triple product, which can also be expressed as \(\vec{a} \cdot (\vec{b} \times \vec{c})\).

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

We start by taking the dot product of \(\vec{\alpha}\) with \(\vec{a}\):

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

### Step 2: Simplify the terms

Using the properties of the dot and cross products:

1. \(\vec{a} \cdot (\vec{a} \times \vec{b}) = 0\) (since the dot product of a vector with a vector perpendicular to it is zero).
2. \(\vec{a} \cdot (\vec{c} \times \vec{a}) = 0\) (for the same reason).

Thus, we simplify:

\[ \vec{\alpha} \cdot \vec{a} = 0 + y(\vec{a} \cdot (\vec{b} \times \vec{c})) + 0 = y[\vec{a}, \vec{b}, \vec{c}] \]

### Step 3: Substitute the value of the scalar triple product

Given that \([\vec{a}, \vec{b}, \vec{c}] = \frac{1}{8}\):

\[ \vec{\alpha} \cdot \vec{a} = y \cdot \frac{1}{8} \]

### Step 4: Repeat for \(\vec{b}\) and \(\vec{c}\)

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

\[ \vec{\alpha} \cdot \vec{b} = x(\vec{b} \cdot (\vec{a} \times \vec{b})) + y(\vec{b} \cdot (\vec{b} \times \vec{c})) + z(\vec{b} \cdot (\vec{c} \times \vec{a})) = 0 + 0 + z[\vec{a}, \vec{b}, \vec{c}] = z \cdot \frac{1}{8} \]

Next, for \(\vec{c}\):

\[ \vec{\alpha} \cdot \vec{c} = x(\vec{c} \cdot (\vec{a} \times \vec{b})) + y(\vec{c} \cdot (\vec{b} \times \vec{c})) + z(\vec{c} \cdot (\vec{c} \times \vec{a})) = x[\vec{a}, \vec{b}, \vec{c}] + 0 + 0 = x \cdot \frac{1}{8} \]

### Step 5: Express \(x\), \(y\), and \(z\)

From the above equations, we can express \(x\), \(y\), and \(z\):

\[ y = 8(\vec{\alpha} \cdot \vec{a}), \quad z = 8(\vec{\alpha} \cdot \vec{b}), \quad x = 8(\vec{\alpha} \cdot \vec{c}) \]

### Step 6: Add \(x\), \(y\), and \(z\)

Now, we can find \(x + y + z\):

\[ x + y + z = 8(\vec{\alpha} \cdot \vec{c}) + 8(\vec{\alpha} \cdot \vec{b}) + 8(\vec{\alpha} \cdot \vec{a}) = 8(\vec{\alpha} \cdot (\vec{a} + \vec{b} + \vec{c})) \]

### Final Result

Thus, the final answer is:

\[ x + y + z = 8(\vec{\alpha} \cdot (\vec{a} + \vec{b} + \vec{c})) \]