If `vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` and `[veca vecb vecc]=(1)/(3)`, then x+y+z is equal to

Options

A. `7vecr.(veca+vecb+vecc)`
B. `vecr.(veca+vecb+vecc)`
C. `5vecr.(veca+vecb+vecc)`
D. `3vecr.(veca+vecb+vecc)`

Answer

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

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

We also know that the scalar triple product \([\vec{a}, \vec{b}, \vec{c}] = \frac{1}{3}\).

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

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

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

Since \(\vec{a} \times \vec{b}\) is perpendicular to \(\vec{a}\), the first term becomes 0:

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

Thus, we have:

\[ \vec{r} \cdot \vec{a} = y(\vec{b} \times \vec{c}) \cdot \vec{a} \]

### Step 2: Use the scalar triple product

The term \((\vec{b} \times \vec{c}) \cdot \vec{a}\) is the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\):

\[ \vec{r} \cdot \vec{a} = y \cdot \frac{1}{3} \]

From this, we can express \(y\):

\[ y = 3(\vec{r} \cdot \vec{a}) \]

### Step 3: Take the dot product with \(\vec{b}\)

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

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

Again, \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\) and \((\vec{b} \times \vec{c}) \cdot \vec{b} = 0\):

\[ \vec{r} \cdot \vec{b} = 0 + 0 + z(\vec{c} \times \vec{a}) \cdot \vec{b} \]

Thus, we have:

\[ \vec{r} \cdot \vec{b} = z(\vec{c} \times \vec{a}) \cdot \vec{b} \]

Using the scalar triple product:

\[ \vec{r} \cdot \vec{b} = z \cdot \frac{1}{3} \]

From this, we can express \(z\):

\[ z = 3(\vec{r} \cdot \vec{b}) \]

### Step 4: Take the dot product with \(\vec{c}\)

Now, we take the dot product of the original equation with \(\vec{c}\):

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

Again, \((\vec{b} \times \vec{c}) \cdot \vec{c} = 0\) and \((\vec{c} \times \vec{a}) \cdot \vec{c} = 0\):

\[ \vec{r} \cdot \vec{c} = x(\vec{a} \times \vec{b}) \cdot \vec{c} \]

Using the scalar triple product:

\[ \vec{r} \cdot \vec{c} = x \cdot \frac{1}{3} \]

From this, we can express \(x\):

\[ x = 3(\vec{r} \cdot \vec{c}) \]

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

Now we can add the three equations we derived:

\[ x + y + z = 3(\vec{r} \cdot \vec{c}) + 3(\vec{r} \cdot \vec{a}) + 3(\vec{r} \cdot \vec{b}) \]

Factoring out the 3:

\[ x + y + z = 3(\vec{r} \cdot \vec{a} + \vec{r} \cdot \vec{b} + \vec{r} \cdot \vec{c}) \]

### Conclusion

Thus, the final result is:

\[ x + y + z = 3(\vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})) \]