Answer
a) $\lt 0,2,2 \gt$ or, $2j+2k$
b)$\lt 0, \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
Work Step by Step
a) The cross product is defined as:
$u \times v=\begin{vmatrix}i&j&k\\m_1&m_2&m_3\\n_1&n_2&n_3\end{vmatrix}=\lt m_2n_3-m_3n_2, m_3n_1-m_1n_3, m_1n_2-m_2b_1 \gt$
Now, $u \times v=\lt (1)(-1)-(-1)(1),(-1)(-1)-(1)(-1), (1)(1) -(1)(-1) \gt=\lt 0,2,2 \gt$
or, $2j+2k$
b) The unit vector is defined as: $\hat{n}=\dfrac{u \times v}{|u \times v|}$
Here,$ |u \times v|=\sqrt {(0)^2+(2)^2+(2)^2}=2\sqrt 2$
Now,
$\hat{n}=\dfrac{u \times v}{|u \times v|}=\dfrac{\lt 0,2,2 \gt}{2\sqrt 2}=\lt 0, \dfrac{\sqrt 2}{2},\dfrac{\sqrt 2}{2} \gt$
