Answer
a) $\lt 1,11,-19 \gt$ or, $i+11j-19k$
b) $\lt \dfrac{1}{\sqrt {483}},\dfrac{11}{\sqrt {483}},\dfrac{-19}{\sqrt {483}} \gt$
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, $u \times v=\lt (5)(-1)-(3)(-2),(3)(3)-(2)(-1), (2)(-2) -(5)(3) \gt=\lt 1,11,-19 \gt$
or, $i+11j-19k$
b) The unit vector is defined as: $\hat{n}=\dfrac{u \times v}{|u \times v|}$
Here,$ |u \times v|=\sqrt {(1)^2+(11)^2+(-19)^2}=\sqrt {483}$
Now,
$\hat{n}=\dfrac{u \times v}{|u \times v|}=\dfrac{\lt 1,11,-19 \gt}{\sqrt {483}}=\lt \dfrac{1}{\sqrt {483}},\dfrac{11}{\sqrt {483}},\dfrac{-19}{\sqrt {483}} \gt$
