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 step by step, we will analyze the vectors and calculate the required values.

### Step 1: Define the vectors
We have three vectors:
- \(\vec{p} = \hat{i} + \hat{j} + \hat{k}\)
- \(\vec{q} = 3\hat{i} - \hat{j} + \hat{k}\)
- \(\vec{r} = \alpha \hat{i} + \beta \hat{j} + \lambda \hat{k}\)

### Step 2: Calculate the cross product \(\vec{p} \times \vec{q}\)
To find the box product \([\vec{p}, \vec{q}, \vec{r}]\) which is maximized, we first need to compute \(\vec{p} \times \vec{q}\).

The determinant for the cross product is given by:
\[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 3 & -1 & 1 \end{vmatrix} \]

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

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2\)
2. \(\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} = (1)(1) - (1)(3) = 1 - 3 = -2\)
3. \(\begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (1)(3) = -1 - 3 = -4\)

Thus, we have:
\[ \vec{p} \times \vec{q} = 2\hat{i} + 2\hat{j} - 4\hat{k} \]

### Step 3: Condition for maximum box product
The box product \([\vec{p}, \vec{q}, \vec{r}]\) can be expressed as:
\[ [\vec{p}, \vec{q}, \vec{r}] = \vec{p} \times \vec{q} \cdot \vec{r} \]
To maximize this, \(\vec{r}\) should be parallel to \(\vec{p} \times \vec{q}\). Therefore, we can write:
\[ \vec{r} = \mu (\vec{p} \times \vec{q}) = \mu (2\hat{i} + 2\hat{j} - 4\hat{k}) \]

### Step 4: Magnitude of \(\vec{r}\)
Given that \(|\vec{r}| = 2\sqrt{6}\), we can express this as:
\[ |\vec{r}| = |\mu| \cdot |\vec{p} \times \vec{q}| \]
Calculating the magnitude of \(\vec{p} \times \vec{q}\):
\[ |\vec{p} \times \vec{q}| = \sqrt{(2)^2 + (2)^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \]

Thus, we have:
\[ |\vec{r}| = |\mu| \cdot 2\sqrt{6} \]
Setting this equal to \(2\sqrt{6}\):
\[ |\mu| \cdot 2\sqrt{6} = 2\sqrt{6} \implies |\mu| = 1 \]
So, \(\mu = 1\) or \(\mu = -1\).

### Step 5: Finding \(\alpha\), \(\beta\), and \(\lambda\)
Using \(\mu = 1\):
\[ \vec{r} = 2\hat{i} + 2\hat{j} - 4\hat{k} \]
Thus, we have:
- \(\alpha = 2\)
- \(\beta = 2\)
- \(\lambda = -4\)

### Step 6: Calculate \(\alpha - \beta - \lambda\)
Now we calculate:
\[ \alpha - \beta - \lambda = 2 - 2 - (-4) = 2 - 2 + 4 = 4 \]

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