If `vecp =3veci +sqrt 3vecj +2veck` and `vecQ=4veci +sqrt3vec j+2.5veck` then, the unit vector in the direction of `vecP times vecQ` is `1/x (sqrt3i+vecj-2sqrt3veck)`. The value of x is

Answer

To solve the problem, we need to find the unit vector in the direction of the cross product of two vectors \(\vec{P}\) and \(\vec{Q}\). Let's go through the steps systematically.

### Step 1: Define the vectors
Given:
\[ \vec{P} = 3\hat{i} + \sqrt{3}\hat{j} + 2\hat{k} \]
\[ \vec{Q} = 4\hat{i} + \sqrt{3}\hat{j} + 2.5\hat{k} \]

### Step 2: Set up the cross product
The cross product \(\vec{P} \times \vec{Q}\) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \(\vec{P}\) and \(\vec{Q}\):

\[ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \vec{P} \times \vec{Q} = \hat{i} \begin{vmatrix} \sqrt{3} & 2 \\ \sqrt{3} & 2.5 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 4 & 2.5 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & \sqrt{3} \\ 4 & \sqrt{3} \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} \sqrt{3} & 2 \\ \sqrt{3} & 2.5 \end{vmatrix} = \sqrt{3} \cdot 2.5 - 2 \cdot \sqrt{3} = 2.5\sqrt{3} - 2\sqrt{3} = 0.5\sqrt{3} \]
2. For \(\hat{j}\):
\[ \begin{vmatrix} 3 & 2 \\ 4 & 2.5 \end{vmatrix} = 3 \cdot 2.5 - 2 \cdot 4 = 7.5 - 8 = -0.5 \]
3. For \(\hat{k}\):
\[ \begin{vmatrix} 3 & \sqrt{3} \\ 4 & \sqrt{3} \end{vmatrix} = 3\sqrt{3} - 4\sqrt{3} = -\sqrt{3} \]

Putting it all together:
\[ \vec{P} \times \vec{Q} = 0.5\sqrt{3} \hat{i} + 0.5 \hat{j} + (-\sqrt{3}) \hat{k} \]
Thus,
\[ \vec{P} \times \vec{Q} = 0.5\sqrt{3} \hat{i} - 0.5 \hat{j} - \sqrt{3} \hat{k} \]

### Step 4: Find the magnitude of the cross product
To find the unit vector, we first need the magnitude of \(\vec{P} \times \vec{Q}\):
\[ |\vec{P} \times \vec{Q}| = \sqrt{(0.5\sqrt{3})^2 + (0.5)^2 + (-\sqrt{3})^2} \]
Calculating:
\[ = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = \sqrt{\frac{3 + 1 + 12}{4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2 \]

### Step 5: Find the unit vector
The unit vector in the direction of \(\vec{P} \times \vec{Q}\) is:
\[ \hat{R} = \frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|} = \frac{0.5\sqrt{3} \hat{i} + 0.5 \hat{j} - \sqrt{3} \hat{k}}{2} \]
This simplifies to:
\[ \hat{R} = \frac{1}{4} \sqrt{3} \hat{i} + \frac{1}{4} \hat{j} - \frac{\sqrt{3}}{2} \hat{k} \]

### Step 6: Compare with given unit vector
We are given that the unit vector is:
\[ \frac{1}{x} (\sqrt{3} \hat{i} + \hat{j} - 2\sqrt{3} \hat{k}) \]
Setting the two expressions equal:
\[ \frac{1}{4} \sqrt{3} \hat{i} + \frac{1}{4} \hat{j} - \frac{\sqrt{3}}{2} \hat{k} = \frac{1}{x} (\sqrt{3} \hat{i} + \hat{j} - 2\sqrt{3} \hat{k}) \]

### Step 7: Solve for \(x\)
From the coefficients of \(\hat{i}\):
\[ \frac{1}{4} \sqrt{3} = \frac{\sqrt{3}}{x} \implies x = 4 \]

Thus, the value of \(x\) is:
\[ \boxed{4} \]