Consider three vectors `vecv_(1)=vecv_(2) -vecv_(3),vecv_(1), vecv_(2) and vecv_(3)` such that If `vecv_(1) = (veca xx hati) xx hati, vecv_(2)=(veca xx hatj) xx hatj and vecv_(3) = (veca xx hatk) xx hatk`, where `veca` is non -zero vector then :-

Options

A. `veca *hatj =0`
B. `veca *hati =0`
C. `veca*hatk=0`
D. `vecv_(1) *vecv_(2)=(veca *hatj)^(2)`

Answer

To solve the given problem involving vectors \( \vec{v_1}, \vec{v_2}, \) and \( \vec{v_3} \), we will follow a systematic approach using vector algebra.

### Step 1: Define the vectors
We are given:
- \( \vec{v_1} = \vec{a} \times \hat{i} \times \hat{i} \)
- \( \vec{v_2} = \vec{a} \times \hat{j} \times \hat{j} \)
- \( \vec{v_3} = \vec{a} \times \hat{k} \times \hat{k} \)

### Step 2: Apply the vector triple product identity
The vector triple product identity states that:
\[ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C} \]

Using this identity, we can simplify \( \vec{v_1}, \vec{v_2}, \) and \( \vec{v_3} \).

### Step 3: Calculate \( \vec{v_1} \)
Using the identity:
\[ \vec{v_1} = \vec{a} \times \hat{i} \times \hat{i} = \vec{a} \cdot \hat{i} \hat{i} - \vec{a} \cdot \hat{i} \hat{i} = 0 \]
This means \( \vec{v_1} \) is a zero vector.

### Step 4: Calculate \( \vec{v_2} \)
Similarly,
\[ \vec{v_2} = \vec{a} \times \hat{j} \times \hat{j} = \vec{a} \cdot \hat{j} \hat{j} - \vec{a} \cdot \hat{j} \hat{j} = 0 \]
Thus, \( \vec{v_2} \) is also a zero vector.

### Step 5: Calculate \( \vec{v_3} \)
Using the same logic,
\[ \vec{v_3} = \vec{a} \times \hat{k} \times \hat{k} = \vec{a} \cdot \hat{k} \hat{k} - \vec{a} \cdot \hat{k} \hat{k} = 0 \]
So, \( \vec{v_3} \) is also a zero vector.

### Step 6: Analyze the relationship
We know from the problem statement that:
\[ \vec{v_1} = \vec{v_2} - \vec{v_3} \]
Since all three vectors are zero vectors, this equation holds true.

### Step 7: Determine the implications for vector \( \vec{a} \)
Since \( \vec{v_1}, \vec{v_2}, \) and \( \vec{v_3} \) are all zero vectors, we can conclude that:
- \( \vec{a} \cdot \hat{i} = 0 \)
- \( \vec{a} \cdot \hat{j} = 0 \)
- \( \vec{a} \cdot \hat{k} = 0 \)

This implies that \( \vec{a} \) must be orthogonal to all unit vectors \( \hat{i}, \hat{j}, \hat{k} \), meaning \( \vec{a} \) is the zero vector.

### Step 8: Check the options
Given the options, we need to check which one is correct. The first option states:
\[ \vec{a} \cdot \hat{j} = 0 \]
This is indeed true since \( \vec{a} \) is orthogonal to \( \hat{j} \).

### Conclusion
The correct answer is that the first option is valid.

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