Let `vecV = 2hati +hatj - hatk and vecW= hati + 3hatk . if vecU` is a unit vector, then the maximum value of the scalar triple product `[ vecU vecV vecW]` is

Options

A. `-1`
B. `sqrt10 + sqrt6`
C. `sqrt59`
D. `sqrt60`

Explanation

To find the maximum value of the scalar triple product \([ \vec{U}, \vec{V}, \vec{W} ]\), we will follow these steps:

### Step 1: Define the vectors
We are given:
\[ \vec{V} = 2\hat{i} + \hat{j} - \hat{k} \]
\[ \vec{W} = \hat{i} + 3\hat{k} \]

### Step 2: Calculate the cross product \(\vec{V} \times \vec{W}\)
To find \(\vec{V} \times \vec{W}\), we will use the determinant of a matrix formed by the unit vectors and the components of \(\vec{V}\) and \(\vec{W}\):
\[ \vec{V} \times \vec{W} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -1 \\ 0 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot 3 - 0 \cdot -1) - \hat{j} (2 \cdot 3 - 1 \cdot -1) + \hat{k} (2 \cdot 0 - 1 \cdot 1) \]
\[ = 3\hat{i} - (6 + 1)\hat{j} - \hat{k} \]
\[ = 3\hat{i} - 7\hat{j} - \hat{k} \]

### Step 3: Calculate the magnitude of \(\vec{V} \times \vec{W}\)
Now, we need to find the magnitude of \(\vec{V} \times \vec{W}\):
\[ |\vec{V} \times \vec{W}| = \sqrt{3^2 + (-7)^2 + (-1)^2} \]
\[ = \sqrt{9 + 49 + 1} = \sqrt{59} \]

### Step 4: Express the scalar triple product
The scalar triple product can be expressed as:
\[ [\vec{U}, \vec{V}, \vec{W}] = \vec{U} \cdot (\vec{V} \times \vec{W}) \]
Using the property of the dot product, we have:
\[ [\vec{U}, \vec{V}, \vec{W}] = |\vec{U}| |\vec{V} \times \vec{W}| \cos \theta \]
Since \(\vec{U}\) is a unit vector, \(|\vec{U}| = 1\):
\[ [\vec{U}, \vec{V}, \vec{W}] = \sqrt{59} \cos \theta \]

### Step 5: Find the maximum value
The maximum value of \(\cos \theta\) is 1 (when \(\theta = 0\)), thus the maximum value of the scalar triple product is:
\[ \text{Maximum value} = \sqrt{59} \cdot 1 = \sqrt{59} \]

### Final Answer
The maximum value of the scalar triple product \([ \vec{U}, \vec{V}, \vec{W} ]\) is \(\sqrt{59}\).

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