If `veca + 2 vecb + 3 vecc = vec0 " then " veca xx vecb + vecb xx vecc + vecc xx veca= `

Options

A. `2 (veca xx vecb)`
B. ` 6( vecb xx vecc)`
C. `3 ( vecc xx veca)`
D. `vec0`

Explanation

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

\[ \vec{a} + 2\vec{b} + 3\vec{c} = \vec{0} \]

We need to find the value of:

\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \]

### Step 1: Rearranging the Given Equation

From the equation \(\vec{a} + 2\vec{b} + 3\vec{c} = \vec{0}\), we can express \(\vec{a}\) in terms of \(\vec{b}\) and \(\vec{c}\):

\[ \vec{a} = -2\vec{b} - 3\vec{c} \]

### Step 2: Cross Product with \(\vec{a}\)

Now, we will take the cross product of \(\vec{a}\) with the entire equation:

\[ \vec{a} \times (\vec{a} + 2\vec{b} + 3\vec{c}) = \vec{a} \times \vec{0} \]

This gives us:

\[ \vec{a} \times \vec{a} + 2\vec{a} \times \vec{b} + 3\vec{a} \times \vec{c} = \vec{0} \]

Since \(\vec{a} \times \vec{a} = \vec{0}\), we simplify this to:

\[ 2\vec{a} \times \vec{b} + 3\vec{a} \times \vec{c} = \vec{0} \]

### Step 3: Expressing \(\vec{a} \times \vec{c}\)

From the equation above, we can express \(\vec{a} \times \vec{c}\):

\[ 3\vec{a} \times \vec{c} = -2\vec{a} \times \vec{b} \]

Dividing through by 3 gives:

\[ \vec{a} \times \vec{c} = -\frac{2}{3}\vec{a} \times \vec{b} \]

### Step 4: Cross Product with \(\vec{c}\)

Next, we take the cross product of the entire original equation with \(\vec{c}\):

\[ \vec{c} \times (\vec{a} + 2\vec{b} + 3\vec{c}) = \vec{c} \times \vec{0} \]

This simplifies to:

\[ \vec{c} \times \vec{a} + 2\vec{c} \times \vec{b} + 3\vec{c} \times \vec{c} = \vec{0} \]

Since \(\vec{c} \times \vec{c} = \vec{0}\), we have:

\[ \vec{c} \times \vec{a} + 2\vec{c} \times \vec{b} = \vec{0} \]

### Step 5: Expressing \(\vec{c} \times \vec{a}\)

From this equation, we can express \(\vec{c} \times \vec{a}\):

\[ \vec{c} \times \vec{a} = -2\vec{c} \times \vec{b} \]

### Step 6: Substitute Back into the Expression

Now we substitute \(\vec{c} \times \vec{a}\) into our original expression:

\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \]

Using \(\vec{c} \times \vec{a} = -2\vec{c} \times \vec{b}\):

\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} - 2\vec{c} \times \vec{b} \]

### Step 7: Combine Like Terms

Now we can combine the terms:

\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} - 2\vec{b} \times \vec{c} = \vec{a} \times \vec{b} - \vec{b} \times \vec{c} \]

### Step 8: Final Result

Thus, we find that:

\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \frac{1}{2} \vec{a} \times \vec{b} \]

This leads us to conclude that:

\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0} \]

### Final Answer

The final answer is:

\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0} \]