Consider the lines `L_1 : r=a+lambdab and L_2 : r=b+mua`, where a and b are non zero and non collinear vectors.
Statement-I `L_1 and L_2` are coplanar and the plane containing these lines passes through origin.
Statement-II `(a-b)cdot(atimesb)=0` and the plane containing `L_1 and L_2` is [r a b]=0 which passe through origin.

Options

A. Statement-I is true, Statement II is also true, Statement-II is the correct explanation of Statement-I.
B. Statement-I is true, Statement-II is also true, Statement-II is not the correct explanation of Statement-I.
C. Statement-I is true, Statement-II is false.
D. Statement-I is false, Statement -II is true.

Answer

To solve the problem, we need to analyze the given lines \( L_1 \) and \( L_2 \) and the statements provided.

### Step 1: Understand the lines
The lines are given as:
- \( L_1: \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \)
- \( L_2: \mathbf{r} = \mathbf{b} + \mu \mathbf{a} \)

Here, \( \mathbf{a} \) and \( \mathbf{b} \) are non-zero and non-collinear vectors. This means they do not lie on the same line.

**Hint:** Identify the direction vectors of the lines and the points they pass through.

### Step 2: Check for coplanarity
To check if the lines \( L_1 \) and \( L_2 \) are coplanar, we can use the condition involving the scalar triple product. The lines are coplanar if:

\[ (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) = 0 \]

**Hint:** Recall that the scalar triple product gives a volume; if it equals zero, the vectors are coplanar.

### Step 3: Analyze the plane containing the lines
The normal vector to the plane containing the lines can be found using the cross product:

\[ \mathbf{n} = \mathbf{a} \times \mathbf{b} \]

The equation of the plane can be expressed as:

\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \]

where \( \mathbf{r_0} \) is a point on the plane. We can choose \( \mathbf{r_0} = \mathbf{a} \) or \( \mathbf{b} \) since both lines pass through these points.

**Hint:** Remember that the plane equation can be derived from the normal vector and a point on the plane.

### Step 4: Substitute to find the plane equation
Substituting \( \mathbf{r_0} = \mathbf{a} \) into the plane equation gives:

\[ (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{r} - \mathbf{a}) = 0 \]

This simplifies to:

\[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{r} = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} \]

Since \( \mathbf{a} \times \mathbf{a} = \mathbf{0} \), we can conclude that the plane indeed passes through the origin if we set \( \mathbf{r} = \mathbf{0} \).

**Hint:** The condition for the plane to pass through the origin is that the right-hand side of the equation equals zero.

### Step 5: Conclusion about the statements
From the analysis:
- **Statement I**: \( L_1 \) and \( L_2 \) are coplanar, and the plane containing these lines passes through the origin. This is true.
- **Statement II**: The condition \( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) = 0 \) is indeed a condition for coplanarity and confirms that the plane passes through the origin. This is also true.

Thus, both statements are true.

**Final Answer:** Both Statement I and Statement II are true.