If `overline(a)=hat(i)+hat(j)+hat(k), overline(b)=2hat(i)+lambdahat(j)+hat(k), overline(c)=hat(i)-hat(j)+4hat(k) and overline(a)*(overline(b)timesoverline(c))=10`, then `lambda ` is equal to

Options

A. `6`
B. `7`
C. `9`
D. `10`

Answer

To solve the problem step by step, we need to find the value of \( \lambda \) given the vectors \( \overline{a} \), \( \overline{b} \), and \( \overline{c} \) and the equation \( \overline{a} \cdot (\overline{b} \times \overline{c}) = 10 \).

### Step 1: Define the vectors
Given:
- \( \overline{a} = \hat{i} + \hat{j} + \hat{k} \)
- \( \overline{b} = 2\hat{i} + \lambda \hat{j} + \hat{k} \)
- \( \overline{c} = \hat{i} - \hat{j} + 4\hat{k} \)

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

\[ \overline{b} \times \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & \lambda & 1 \\ 1 & -1 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ \overline{b} \times \overline{c} = \hat{i} \begin{vmatrix} \lambda & 1 \\ -1 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & \lambda \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} \lambda & 1 \\ -1 & 4 \end{vmatrix} = 4\lambda + 1 \)

2. \( \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = 8 - 1 = 7 \)

3. \( \begin{vmatrix} 2 & \lambda \\ 1 & -1 \end{vmatrix} = -2 - \lambda \)

Putting it all together:

\[ \overline{b} \times \overline{c} = (4\lambda + 1) \hat{i} - 7 \hat{j} + (-2 - \lambda) \hat{k} \]

### Step 4: Calculate the dot product \( \overline{a} \cdot (\overline{b} \times \overline{c}) \)
Now we compute the dot product:

\[ \overline{a} \cdot (\overline{b} \times \overline{c}) = (1\hat{i} + 1\hat{j} + 1\hat{k}) \cdot ((4\lambda + 1)\hat{i} - 7\hat{j} + (-2 - \lambda)\hat{k}) \]

Calculating the dot product:

\[ = 1(4\lambda + 1) + 1(-7) + 1(-2 - \lambda) \]

Expanding this gives:

\[ = 4\lambda + 1 - 7 - 2 - \lambda = 3\lambda - 8 \]

### Step 5: Set the equation equal to 10
We set the equation equal to 10:

\[ 3\lambda - 8 = 10 \]

### Step 6: Solve for \( \lambda \)
Now, solve for \( \lambda \):

\[ 3\lambda = 10 + 8 \]
\[ 3\lambda = 18 \]
\[ \lambda = \frac{18}{3} = 6 \]

### Final Answer
Thus, the value of \( \lambda \) is \( 6 \).