If the vectors `hati-hatj, hatj+hatk and veca` form a triangle then `veca` may be (A) `-hati-hatk` (B) `hati-2hatj-hatk` (C) `2hati+hatj+hatjk` (D) hati+hatk`

Options

A. `-hati-hatk`
B. `hati-2hatj-hatk`
C. `2hatj+hatj+hatk`
D. `hati+hatk`

Explanation

To determine which of the given options for the vector \( \vec{a} \) allows the vectors \( \hat{i} - \hat{j} \), \( \hat{j} + \hat{k} \), and \( \vec{a} \) to form a triangle, we can use the property that for three vectors to form a triangle, one vector must be expressible as the sum of the other two.

Let:
- \( \vec{x} = \hat{i} - \hat{j} \)
- \( \vec{y} = \hat{j} + \hat{k} \)
- \( \vec{z} = \vec{a} \)

We need to check the following conditions:
1. \( \vec{x} = \vec{y} + \vec{z} \)
2. \( \vec{y} = \vec{z} + \vec{x} \)
3. \( \vec{z} = \vec{x} + \vec{y} \)

### Step 1: Check \( \vec{x} = \vec{y} + \vec{z} \)

Substituting the vectors:
\[ \hat{i} - \hat{j} = (\hat{j} + \hat{k}) + \vec{a} \]
Rearranging gives:
\[ \vec{a} = \hat{i} - \hat{j} - \hat{j} - \hat{k} = \hat{i} - 2\hat{j} - \hat{k} \]

### Step 2: Check \( \vec{y} = \vec{z} + \vec{x} \)

Substituting the vectors:
\[ \hat{j} + \hat{k} = \vec{a} + (\hat{i} - \hat{j}) \]
Rearranging gives:
\[ \vec{a} = \hat{j} + \hat{k} - \hat{i} + \hat{j} = -\hat{i} + 2\hat{j} + \hat{k} \]

### Step 3: Check \( \vec{z} = \vec{x} + \vec{y} \)

Substituting the vectors:
\[ \vec{a} = (\hat{i} - \hat{j}) + (\hat{j} + \hat{k}) \]
This simplifies to:
\[ \vec{a} = \hat{i} + \hat{k} \]

### Summary of Results

From the three checks, we have:
1. \( \vec{a} = \hat{i} - 2\hat{j} - \hat{k} \) from the first condition.
2. \( \vec{a} = -\hat{i} + 2\hat{j} + \hat{k} \) from the second condition.
3. \( \vec{a} = \hat{i} + \hat{k} \) from the third condition.

### Conclusion

The possible vectors \( \vec{a} \) that satisfy the conditions for forming a triangle with the given vectors are:
- \( \hat{i} - 2\hat{j} - \hat{k} \) (Option B)
- \( \hat{i} + \hat{k} \) (Option D)

### Final Answer

The correct options for \( \vec{a} \) are:
- (B) \( \hat{i} - 2\hat{j} - \hat{k} \)
- (D) \( \hat{i} + \hat{k} \)