`a =2i-j+k, b=i+2j-3k, c=3i+lambdaj+5k` and if these vectors be coplanar, then `lambda` is

Options

A. 4
B. 6
C. `-4`
D. 2

Answer

To determine the value of \( \lambda \) such that the vectors \( \mathbf{a} = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \), \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \), and \( \mathbf{c} = 3\mathbf{i} + \lambda \mathbf{j} + 5\mathbf{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors must equal zero.

### Step-by-Step Solution:

1. **Understanding Coplanarity**:
Vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are coplanar if the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \).

2. **Calculate \( \mathbf{b} \times \mathbf{c} \)**:
We will first calculate the cross product \( \mathbf{b} \times \mathbf{c} \). The cross product can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors.

\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix} \]

3. **Expanding the Determinant**:
We will expand this determinant using the first row:

\[ \mathbf{b} \times \mathbf{c} = \mathbf{i} \begin{vmatrix} 2 & -3 \\ \lambda & 5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 3 & \lambda \end{vmatrix} \]

- For \( \mathbf{i} \):
\[ = \mathbf{i} (2 \cdot 5 - (-3) \cdot \lambda) = \mathbf{i} (10 + 3\lambda) \]
- For \( \mathbf{j} \):
\[ = -\mathbf{j} (1 \cdot 5 - (-3) \cdot 3) = -\mathbf{j} (5 + 9) = -14\mathbf{j} \]
- For \( \mathbf{k} \):
\[ = \mathbf{k} (1 \cdot \lambda - 2 \cdot 3) = \mathbf{k} (\lambda - 6) \]

Combining these, we get:
\[ \mathbf{b} \times \mathbf{c} = (10 + 3\lambda) \mathbf{i} - 14 \mathbf{j} + (\lambda - 6) \mathbf{k} \]

4. **Calculate \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \)**:
Now we compute the dot product of \( \mathbf{a} \) with \( \mathbf{b} \times \mathbf{c} \):

\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (2\mathbf{i} - \mathbf{j} + \mathbf{k}) \cdot ((10 + 3\lambda) \mathbf{i} - 14 \mathbf{j} + (\lambda - 6) \mathbf{k}) \]

Expanding this:
\[ = 2(10 + 3\lambda) - 1(-14) + 1(\lambda - 6) \]
\[ = 20 + 6\lambda + 14 + \lambda - 6 \]
\[ = 28 + 7\lambda \]

5. **Setting the Scalar Triple Product to Zero**:
For coplanarity, we set the scalar triple product equal to zero:
\[ 28 + 7\lambda = 0 \]

6. **Solving for \( \lambda \)**:
\[ 7\lambda = -28 \]
\[ \lambda = -4 \]

### Final Answer:
The value of \( \lambda \) is \( -4 \).