If `vec(a)= hat(i) + hat(j) + hat(k) and vec(b)= hat(j)-hat(k)`, then find `vec(c )` such that `vec(a ) xx vec(c )= vec(b) and vec(a).vec(c )=3`.

Answer

To solve the problem, we need to find the vector \(\vec{c}\) such that:

1. \(\vec{a} \times \vec{c} = \vec{b}\)
2. \(\vec{a} \cdot \vec{c} = 3\)

Given:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{j} - \hat{k} \]

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

We will use the vector triple product identity:
\[ \vec{a} \times \vec{c} = \vec{b} \implies \vec{a} \times \vec{c} = \vec{b} \]

First, we need to compute \(\vec{a} \times \vec{b}\):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \]

Calculating the minors:
\[ = \hat{i} (1 \cdot (-1) - 1 \cdot 1) - \hat{j} (1 \cdot (-1) - 0 \cdot 1) + \hat{k} (1 \cdot 1 - 0 \cdot 1) \]
\[ = \hat{i} (-1 - 1) - \hat{j} (-1) + \hat{k} (1) \]
\[ = -2\hat{i} + \hat{j} + \hat{k} \]

### Step 2: Set up the equation using the vector triple product identity

Using the identity:
\[ \vec{a} \times \vec{c} = \vec{b} \implies \vec{a} \times \vec{c} = \hat{j} - \hat{k} \]

### Step 3: Use the dot product condition

Now, we also have:
\[ \vec{a} \cdot \vec{c} = 3 \]
Substituting \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\):
\[ (\hat{i} + \hat{j} + \hat{k}) \cdot \vec{c} = 3 \]

Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\):
\[ 1 \cdot x + 1 \cdot y + 1 \cdot z = 3 \implies x + y + z = 3 \]

### Step 4: Solve the equations

From the cross product, we have:
\[ \vec{a} \times \vec{c} = -2\hat{i} + \hat{j} + \hat{k} \]
This gives us a system of equations. We can express \(\vec{c}\) in terms of \(x\), \(y\), and \(z\).

Using the identity:
\[ \vec{a} \cdot \vec{c} = 3 \implies x + y + z = 3 \]

From the cross product, we can also derive:
\[ \vec{a} \cdot \vec{c} = 3 \implies 3 = 3x + 3y + 3z \]

Now, substituting \(x + y + z = 3\) into the equation we derived from the cross product gives us:
\[ -3\vec{c} = -2\hat{i} + \hat{j} + \hat{k} \implies \vec{c} = \frac{-2\hat{i} + \hat{j} + \hat{k}}{-3} \]
\[ = \frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} - \frac{1}{3}\hat{k} \]

### Final Result

Thus, the vector \(\vec{c}\) is:
\[ \vec{c} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{1}{3}\hat{k} \]