Let a, b and c are non-zero vectors such that they are not orthogonal pairwise and such that `V_1 =atimes(btimesc) and V_2=(atimesb)timesc, `are collinear then which of the following holds goods?

Option1. a and b are orthogonal
Option 2. a and c are collinear
Option 3. b and c are orthogonal
Option 4. `b=lambda(atimesc)` when `lambda` is a scalar

Options

A. a and b are orthogonal
B. a and c are collinear
C. b and c are orthogonal
D. `b=lambda(atimesc)` when `lambda` is a scalar

Answer

To solve the problem step by step, we need to analyze the given vectors and their relationships. We have two vectors \( V_1 \) and \( V_2 \) defined as follows:

\[ V_1 = \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \]
\[ V_2 = (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} \]

We are given that \( V_1 \) and \( V_2 \) are collinear. This means there exists a scalar \( \lambda \) such that:

\[ V_1 = \lambda V_2 \]

### Step 1: Apply the Vector Triple Product Identity

Using the vector triple product identity, we can simplify both \( V_1 \) and \( V_2 \).

For \( V_1 \):
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]

For \( V_2 \):
\[ (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{c} \cdot \mathbf{a}) \mathbf{b} - (\mathbf{c} \cdot \mathbf{b}) \mathbf{a} \]

### Step 2: Set the Equations Equal

Since \( V_1 = \lambda V_2 \), we can write:
\[ (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} = \lambda \left( (\mathbf{c} \cdot \mathbf{a}) \mathbf{b} - (\mathbf{c} \cdot \mathbf{b}) \mathbf{a} \right) \]

### Step 3: Compare Coefficients

Now, we can compare the coefficients of \( \mathbf{b} \) and \( \mathbf{c} \) on both sides of the equation.

From the coefficients of \( \mathbf{b} \):
\[ \mathbf{a} \cdot \mathbf{c} = \lambda (\mathbf{c} \cdot \mathbf{a}) \]
This simplifies to:
\[ \mathbf{a} \cdot \mathbf{c} = \lambda (\mathbf{a} \cdot \mathbf{c}) \]

From the coefficients of \( \mathbf{c} \):
\[ -(\mathbf{a} \cdot \mathbf{b}) = -\lambda (\mathbf{c} \cdot \mathbf{b}) \]
This simplifies to:
\[ \mathbf{a} \cdot \mathbf{b} = \lambda (\mathbf{c} \cdot \mathbf{b}) \]

### Step 4: Analyze the Results

From the first equation, if \( \mathbf{a} \cdot \mathbf{c} \neq 0 \), we can divide both sides by \( \mathbf{a} \cdot \mathbf{c} \) to find \( \lambda = 1 \).

From the second equation, we can express \( \mathbf{b} \) in terms of \( \mathbf{a} \) and \( \mathbf{c} \):
\[ \mathbf{b} = \lambda \frac{\mathbf{c} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{b}} \mathbf{c} \]
This indicates that \( \mathbf{b} \) is a scalar multiple of \( \mathbf{c} \).

### Conclusion

Now we can evaluate the options:

1. **Option 1:** \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal - **False** (given they are not orthogonal pairwise).
2. **Option 2:** \( \mathbf{a} \) and \( \mathbf{c} \) are collinear - **True** (as derived).
3. **Option 3:** \( \mathbf{b} \) and \( \mathbf{c} \) are orthogonal - **False** (not supported by the analysis).
4. **Option 4:** \( \mathbf{b} = \lambda (\mathbf{a} \times \mathbf{c}) \) when \( \lambda \) is a scalar - **True** (as derived).

Thus, the correct options are **Option 2** and **Option 4**.