If `vec(a)=hat(i)+hat(j)+hat(k), vec(b)=hat(i)-hat(j)+hat(k) and vec(c)=hat(i)+hat(j)-hat(k)`, then what is `vec(a)xx(vec(b)+vec(c))+vec(b)xx(vec(c)+vec(a))+vec(c)xx(vec(a)+vec(b))` equal to?

Options

A. `2hat(i)+3hat(j)-hat(k)`
B. `2hat(i)-3hat(j)-hat(k)`
C. `3hat(i)+hat(j)+hat(k)`
D. `vec(0)`

Explanation

To solve the problem, we need to evaluate the expression:

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

Given:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]
\[ \vec{c} = \hat{i} + \hat{j} - \hat{k} \]

### Step 1: Calculate \(\vec{b} + \vec{c}\)

\[ \vec{b} + \vec{c} = (\hat{i} - \hat{j} + \hat{k}) + (\hat{i} + \hat{j} - \hat{k}) = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i} \]

### Step 2: Calculate \(\vec{c} + \vec{a}\)

\[ \vec{c} + \vec{a} = (\hat{i} + \hat{j} - \hat{k}) + (\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} + 2\hat{j} + 0\hat{k} = 2\hat{i} + 2\hat{j} \]

### Step 3: Calculate \(\vec{a} + \vec{b}\)

\[ \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} - \hat{j} + \hat{k}) = 2\hat{i} + 0\hat{j} + 2\hat{k} = 2\hat{i} + 2\hat{k} \]

### Step 4: Calculate \(\vec{a} \times (\vec{b} + \vec{c})\)

\[ \vec{a} \times (2\hat{i}) = (1\hat{i} + 1\hat{j} + 1\hat{k}) \times (2\hat{i}) = 2(\hat{j} \times \hat{i} + \hat{k} \times \hat{i}) = 2(-\hat{k} + \hat{j}) = 2\hat{j} - 2\hat{k} \]

### Step 5: Calculate \(\vec{b} \times (\vec{c} + \vec{a})\)

\[ \vec{b} \times (2\hat{i} + 2\hat{j}) = (\hat{i} - \hat{j} + \hat{k}) \times (2\hat{i} + 2\hat{j}) = 2(\hat{i} \times \hat{i} + \hat{j} \times \hat{i}) - 2(\hat{j} \times \hat{j} + \hat{k} \times \hat{i}) = 2(0 + \hat{k}) - 2(0 - \hat{j}) = 2\hat{k} + 2\hat{j} \]

### Step 6: Calculate \(\vec{c} \times (\vec{a} + \vec{b})\)

\[ \vec{c} \times (2\hat{i} + 2\hat{k}) = (\hat{i} + \hat{j} - \hat{k}) \times (2\hat{i} + 2\hat{k}) = 2(\hat{i} \times \hat{i} + \hat{j} \times \hat{i}) - 2(\hat{k} \times \hat{i}) = 2(0 + \hat{k}) - 2(-\hat{j}) = 2\hat{k} + 2\hat{j} \]

### Step 7: Combine all the results

Now we sum up all the cross products calculated:

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

\[ = (2\hat{j} - 2\hat{k}) + (2\hat{k} + 2\hat{j}) + (2\hat{k} + 2\hat{j}) \]

\[ = 2\hat{j} - 2\hat{k} + 2\hat{k} + 2\hat{j} + 2\hat{k} + 2\hat{j} \]

\[ = (2\hat{j} + 2\hat{j} + 2\hat{j}) + (-2\hat{k} + 2\hat{k}) = 6\hat{j} + 0\hat{k} = 6\hat{j} \]

### Final Result

Thus, the final answer is:

\[ \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) = 6\hat{j} \]