Answer
a) $\lt 14,7,0 \gt$ or, $14i+7j$
b) $ \lt \dfrac{14}{\sqrt {245}},\dfrac{7}{\sqrt {245}},0 \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, $u \times v=\lt (-1)(-6)-(\dfrac{2}{3})(-12),(\dfrac{2}{3})(6)-(\dfrac{1}{2})(-6), (\dfrac{1}{2})(-12) -(-1)(6) \gt=\lt 14,7,0 \gt$
or, $14i+7j$
b) The unit vector is defined as: $\hat{n}=\dfrac{u \times v}{|u \times v|}$
Here,$ |u \times v|=\sqrt {(14)^2+(7)^2}=\sqrt {245}$
Now,
$\hat{n}=\dfrac{u \times v}{|u \times v|}=\dfrac{\lt 14,7,0 \gt}{\sqrt {245}}=\lt \dfrac{14}{\sqrt {245}},\dfrac{7}{\sqrt {245}},0 \gt$
