If `vecr=x_(1)(vecaxx vecb) + x_(2) (vecb xxveca) + x_(3)(vecc xxvecd) and 4[veca vecb vecc] =1 " then " x_(1) + x_(2) + x_(3)` is equal to

Options

A. `1/2vecr.(veca + vecb + vecc)`
B. `1/4vecr.(veca + vecb + vecc)`
C. `2vecr.(veca + vecb + vecc)`
D. `4vecr.(veca + vecb + vecc)`

Explanation

To solve the problem step-by-step, we need to find the values of \( x_1 \), \( x_2 \), and \( x_3 \) from the given vector equation and then sum them up.

### Step 1: Understand the given vector equation
We have the vector \( \vec{r} = x_1 (\vec{a} \times \vec{b}) + x_2 (\vec{b} \times \vec{a}) + x_3 (\vec{c} \times \vec{d}) \).

### Step 2: Use the property of dot product and cross product
To find \( x_1 \), \( x_2 \), and \( x_3 \), we can use the property that if two vectors are the same in a cross product, the result is zero.

### Step 3: Calculate \( x_2 \)
We can find \( x_2 \) by taking the dot product of \( \vec{r} \) with \( \vec{a} \):
\[ \vec{r} \cdot \vec{a} = x_1 (\vec{a} \times \vec{b}) \cdot \vec{a} + x_2 (\vec{b} \times \vec{a}) \cdot \vec{a} + x_3 (\vec{c} \times \vec{d}) \cdot \vec{a} \]
The first term becomes zero because \( \vec{a} \times \vec{b} \cdot \vec{a} = 0 \), and the second term also becomes zero because \( \vec{b} \times \vec{a} \cdot \vec{a} = 0 \). Thus, we have:
\[ \vec{r} \cdot \vec{a} = x_3 (\vec{c} \times \vec{d}) \cdot \vec{a} \]

### Step 4: Calculate \( x_3 \)
Now, we can find \( x_3 \) by taking the dot product of \( \vec{r} \) with \( \vec{b} \):
\[ \vec{r} \cdot \vec{b} = x_1 (\vec{a} \times \vec{b}) \cdot \vec{b} + x_2 (\vec{b} \times \vec{a}) \cdot \vec{b} + x_3 (\vec{c} \times \vec{d}) \cdot \vec{b} \]
Again, the first two terms become zero. Thus, we have:
\[ \vec{r} \cdot \vec{b} = x_3 (\vec{c} \times \vec{d}) \cdot \vec{b} \]

### Step 5: Calculate \( x_1 \)
Next, we find \( x_1 \) by taking the dot product of \( \vec{r} \) with \( \vec{c} \):
\[ \vec{r} \cdot \vec{c} = x_1 (\vec{a} \times \vec{b}) \cdot \vec{c} + x_2 (\vec{b} \times \vec{a}) \cdot \vec{c} + x_3 (\vec{c} \times \vec{d}) \cdot \vec{c} \]
The last term becomes zero. Thus, we have:
\[ \vec{r} \cdot \vec{c} = x_1 (\vec{a} \times \vec{b}) \cdot \vec{c} + x_2 (\vec{b} \times \vec{a}) \cdot \vec{c} \]

### Step 6: Use the determinant condition
Given that \( 4[\vec{a}, \vec{b}, \vec{c}] = 1 \), we can express \( [\vec{a}, \vec{b}, \vec{c}] \) as \( \frac{1}{4} \).

### Step 7: Substitute values
From the above equations, we can express:
- \( x_1 = 4 \frac{\vec{r} \cdot \vec{c}}{[\vec{a}, \vec{b}, \vec{c}]} = 16 \vec{r} \cdot \vec{c} \)
- \( x_2 = 4 \frac{\vec{r} \cdot \vec{a}}{[\vec{b}, \vec{c}, \vec{a}]} = 16 \vec{r} \cdot \vec{a} \)
- \( x_3 = 4 \frac{\vec{r} \cdot \vec{b}}{[\vec{c}, \vec{a}, \vec{b}]} = 16 \vec{r} \cdot \vec{b} \)

### Step 8: Find \( x_1 + x_2 + x_3 \)
Now, we can add these values:
\[ x_1 + x_2 + x_3 = 16 (\vec{r} \cdot \vec{c} + \vec{r} \cdot \vec{a} + \vec{r} \cdot \vec{b}) \]

### Step 9: Final expression
Thus, the final value of \( x_1 + x_2 + x_3 \) is:
\[ x_1 + x_2 + x_3 = 4 \cdot \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) \]

### Summary
The answer is \( 4 \cdot \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) \).