If `veca = hati + hatj + hatk, vecc = hatj - hatk` are given vectors, then find a vector `vecb` satisfying the equation `veca xx vecb = vecc and veca. vecb =3.`

Answer

To solve the problem, we need to find a vector \(\vec{b}\) that satisfies the equations \(\vec{a} \times \vec{b} = \vec{c}\) and \(\vec{a} \cdot \vec{b} = 3\). Given the vectors:

\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{c} = \hat{j} - \hat{k} \]

### Step 1: Calculate \(\vec{a} \times \vec{c}\)

To find \(\vec{b}\), we first compute the cross product \(\vec{a} \times \vec{c}\).

\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{a} \times \vec{c} = \hat{i} \left(1 \cdot (-1) - 1 \cdot 1\right) - \hat{j} \left(1 \cdot (-1) - 1 \cdot 0\right) + \hat{k} \left(1 \cdot 1 - 1 \cdot 0\right) \]

This simplifies to:

\[ \vec{a} \times \vec{c} = \hat{i}(-1 - 1) - \hat{j}(-1) + \hat{k}(1) \]
\[ = -2\hat{i} + \hat{j} + \hat{k} \]

### Step 2: Use the vector triple product identity

We use the vector triple product identity:

\[ \vec{a} \times \vec{b} = \vec{c} \implies \vec{b} = \frac{1}{|\vec{a}|^2}(\vec{c} \cdot \vec{a})\vec{a} - \frac{1}{|\vec{a}|^2}(\vec{a} \cdot \vec{b})\vec{c} \]

### Step 3: Calculate \(|\vec{a}|^2\)

First, we compute the magnitude squared of \(\vec{a}\):

\[ |\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3 \]

### Step 4: Substitute values into the equation

Substituting \(|\vec{a}|^2\) and the known values into the equation:

\[ \vec{a} \cdot \vec{b} = 3 \]

We can express \(\vec{b}\) in terms of its components:

\[ \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 5: Set up the equations

From \(\vec{a} \cdot \vec{b} = 3\):

\[ 1 \cdot x + 1 \cdot y + 1 \cdot z = 3 \implies x + y + z = 3 \]

From \(\vec{a} \times \vec{b} = \vec{c}\):

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]

Calculating this determinant gives:

\[ = \hat{i}(1z - 1y) - \hat{j}(1z - 1x) + \hat{k}(1y - 1x) \]
\[ = (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} \]

Setting this equal to \(\vec{c} = \hat{j} - \hat{k}\):

\[ (z - y) = 0 \quad \text{(1)} \]
\[ -(z - x) = 1 \quad \text{(2)} \]
\[ (y - x) = -1 \quad \text{(3)} \]

### Step 6: Solve the equations

From equation (1), we have:

\[ z = y \]

Substituting \(z = y\) into equation (2):

\[ -(y - x) = 1 \implies y - x = -1 \implies x = y + 1 \]

Substituting \(x = y + 1\) into equation (3):

\[ y - (y + 1) = -1 \implies -1 = -1 \quad \text{(True)} \]

### Step 7: Substitute back to find \(x, y, z\)

Using \(x + y + z = 3\):

\[ (y + 1) + y + y = 3 \implies 3y + 1 = 3 \implies 3y = 2 \implies y = \frac{2}{3} \]

Thus,

\[ z = y = \frac{2}{3}, \quad x = y + 1 = \frac{2}{3} + 1 = \frac{5}{3} \]

### Final Result

The vector \(\vec{b}\) is:

\[ \vec{b} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \]