## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

a) $v \times w =−i−6j−k$ b) $w \times v=i+6j+k$ c) $v \times v = 0$ d) $w \times w =0$
Suppose that the two vectors can be represented as: $v=v_1i+v_2j+v_3k$ and $w=w_1i+w_2j+w_3k$ Then the cross product of such vectors can be obtained in the form of the determinate as: $v \times w=\begin{vmatrix} i & j & k \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \\ \end{vmatrix}=(v_2w_3-v_3w_2)i-(v_1w_3-v_3w_1)j+(v_1w_2-v_2w_1)k$ a) Thus, we have the cross product: $v \times w =\begin{vmatrix} i & j & k \\ 3 & 1 &3 \\ 1 & 0 & -1 \\ \end{vmatrix}\\=i(1⋅(−1)−3⋅0)−j(3⋅(−1)−3⋅1)+k(3⋅0−1⋅1)\\=−i−6j−k$ b) The cross product is not commutative. So we can write as: $v \times w= -w \times v$ Thus, $w \times v=i+6j+k$ c) We know that for the two mutually parallel vectors, we have: $v \times v = 0$ d) We know that for the two mutually parallel vectors, we have: $w \times w =0$