If \(\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{j} - \hat{k}\) and \(\vec{a} \times \vec{c} = \vec{b}, \vec{a} \cdot \vec{c} = 3\), then \(\vec{c}\) is

Given \(\vec{a} \times \vec{c} = \vec{b}\)

Taking the cross product with \(\vec{a}\) on both sides:

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

Using the vector triple product expansion:

\((\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c} = \vec{a} \times \vec{b}\)

We are given \(\vec{a} \cdot \vec{c} = 3\). Also, \(\vec{a} \cdot \vec{a} = (1)^2 + (1)^2 + (1)^2 = 3\).

Substituting these values:

\(3\vec{a} - 3\vec{c} = \vec{a} \times \vec{b}\)

\(3\vec{c} = 3\vec{a} - (\vec{a} \times \vec{b})\)

\(\vec{c} = \vec{a} - \dfrac{1}{3}(\vec{a} \times \vec{b})\)

Now, calculating \(\vec{a} \times \vec{b}\):

\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(-1 - 0) + \hat{k}(1 - 0) = -2\hat{i} + \hat{j} + \hat{k}\)

Substituting \(\vec{a}\) and \(\vec{a} \times \vec{b}\) into the equation for \(\vec{c}\):

\(\vec{c} = (\hat{i} + \hat{j} + \hat{k}) - \dfrac{1}{3}(-2\hat{i} + \hat{j} + \hat{k})\)

\(\vec{c} = \left(1 + \dfrac{2}{3}\right)\hat{i} + \left(1 - \dfrac{1}{3}\right)\hat{j} + \left(1 - \dfrac{1}{3}\right)\hat{k}\)

\(\vec{c} = \dfrac{5}{3}\hat{i} + \dfrac{2}{3}\hat{j} + \dfrac{2}{3}\hat{k}\)

Answer: \(\dfrac{5}{3}\hat{i} + \dfrac{2}{3}\hat{j} + \dfrac{2}{3}\hat{k}\)