Consider three vectors `vecp=hati+hatj+hatk, vecq=3hati-hatj+hatk and vecr=alpha hati+betahatj+lambdahatk, AA alpha, beta, lambda in R`. If `[(vecp,vecq,vecr)]` is maximum and `[vecr]=2sqrt6`, then the value of `alpha-beta-lambda` is equal to

Options

A. 8
B. 4
C. 0
D. `-4`

Answer

To solve the problem, we need to find the value of \( \alpha - \beta - \lambda \) given the vectors \( \vec{p} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{q} = 3\hat{i} - \hat{j} + \hat{k} \), and \( \vec{r} = \alpha \hat{i} + \beta \hat{j} + \lambda \hat{k} \). We are also given that the scalar triple product \( [\vec{p}, \vec{q}, \vec{r}] \) is maximum and \( |\vec{r}| = 2\sqrt{6} \).

### Step 1: Calculate the cross product \( \vec{p} \times \vec{q} \)

The scalar triple product can be expressed as:

\[ [\vec{p}, \vec{q}, \vec{r}] = \vec{p} \cdot (\vec{q} \times \vec{r}) \]

To maximize this, we need \( \vec{r} \) to be in the direction of \( \vec{p} \times \vec{q} \). First, we compute \( \vec{p} \times \vec{q} \):

\[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 3 & -1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = 1 \cdot 1 - (-1) \cdot 1 = 1 + 1 = 2 \)
2. \( \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 - 3 \cdot 1 = 1 - 3 = -2 \)
3. \( \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} = 1 \cdot (-1) - 3 \cdot 1 = -1 - 3 = -4 \)

Putting it all together:

\[ \vec{p} \times \vec{q} = 2\hat{i} + 2\hat{j} - 4\hat{k} \]

### Step 2: Find the magnitude of \( \vec{p} \times \vec{q} \)

Now we calculate the magnitude of \( \vec{p} \times \vec{q} \):

\[ |\vec{p} \times \vec{q}| = \sqrt{2^2 + 2^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \]

### Step 3: Express \( \vec{r} \)

Since \( \vec{r} \) must be in the direction of \( \vec{p} \times \vec{q} \), we can express \( \vec{r} \) as:

\[ \vec{r} = k \cdot \left( \vec{p} \times \vec{q} \right) = k(2\hat{i} + 2\hat{j} - 4\hat{k}) \]

Given that \( |\vec{r}| = 2\sqrt{6} \), we find \( k \):

\[ |\vec{r}| = |k| \cdot |\vec{p} \times \vec{q}| = |k| \cdot 2\sqrt{6} \]

Setting this equal to \( 2\sqrt{6} \):

\[ |k| \cdot 2\sqrt{6} = 2\sqrt{6} \implies |k| = 1 \]

Thus, \( k = 1 \) or \( k = -1 \). We can choose \( k = 1 \) for simplicity:

\[ \vec{r} = 2\hat{i} + 2\hat{j} - 4\hat{k} \]

### Step 4: Compare coefficients to find \( \alpha, \beta, \lambda \)

From \( \vec{r} = \alpha \hat{i} + \beta \hat{j} + \lambda \hat{k} \):

- \( \alpha = 2 \)
- \( \beta = 2 \)
- \( \lambda = -4 \)

### Step 5: Calculate \( \alpha - \beta - \lambda \)

Now we compute:

\[ \alpha - \beta - \lambda = 2 - 2 - (-4) = 2 - 2 + 4 = 4 \]

### Final Answer

Thus, the value of \( \alpha - \beta - \lambda \) is:

\[ \boxed{4} \]