What is the value of `lamda` for which `(lamda hat(i) + hat(j)- hat(k)) xx (3 hat(i) -2hat(j) + 4hat(k)) = (2hat(i) - 11 hat(j) - 7hat(k))` ?

Options

A. 2
B. `-2`
C. 1
D. 7

Answer

To solve the problem, we need to find the value of \( \lambda \) such that:

\[ (\lambda \hat{i} + \hat{j} - \hat{k}) \times (3 \hat{i} - 2 \hat{j} + 4 \hat{k}) = (2 \hat{i} - 11 \hat{j} - 7 \hat{k}) \]

### Step 1: Write down the vectors
Let:
\[ \mathbf{A} = \lambda \hat{i} + \hat{j} - \hat{k} \]
\[ \mathbf{B} = 3 \hat{i} - 2 \hat{j} + 4 \hat{k} \]

### Step 2: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)
To find the cross product, we can use the determinant of a matrix formed by the unit vectors and the components of the vectors:

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 1 & -1 \\ 3 & -2 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we expand it as follows:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 1 & -1 \\ -2 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} \lambda & -1 \\ 3 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} \lambda & 1 \\ 3 & -2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & -1 \\ -2 & 4 \end{vmatrix} = (1)(4) - (-1)(-2) = 4 - 2 = 2 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} \lambda & -1 \\ 3 & 4 \end{vmatrix} = (\lambda)(4) - (-1)(3) = 4\lambda + 3 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} \lambda & 1 \\ 3 & -2 \end{vmatrix} = (\lambda)(-2) - (1)(3) = -2\lambda - 3 \]

Putting it all together, we have:
\[ \mathbf{A} \times \mathbf{B} = 2 \hat{i} - (4\lambda + 3) \hat{j} + (-2\lambda - 3) \hat{k} \]

### Step 4: Set the cross product equal to the given vector
Now we set the result equal to the vector given in the problem:
\[ 2 \hat{i} - (4\lambda + 3) \hat{j} + (-2\lambda - 3) \hat{k} = 2 \hat{i} - 11 \hat{j} - 7 \hat{k} \]

### Step 5: Equate the coefficients
From the equation, we can equate the coefficients of \( \hat{j} \) and \( \hat{k} \):

1. For \( \hat{j} \):
\[ -(4\lambda + 3) = -11 \implies 4\lambda + 3 = 11 \]

2. For \( \hat{k} \):
\[ -2\lambda - 3 = -7 \implies 2\lambda + 3 = 7 \]

### Step 6: Solve for \( \lambda \)
From the first equation:
\[ 4\lambda = 11 - 3 = 8 \implies \lambda = \frac{8}{4} = 2 \]

From the second equation:
\[ 2\lambda = 7 - 3 = 4 \implies \lambda = \frac{4}{2} = 2 \]

Both equations give the same value of \( \lambda \).

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 2 \]