Question 3

Write both vectors as column vectors and use the cross product formula from the formula book.

$$\boldsymbol{a}\times\boldsymbol{b}=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\times\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$$

Here $\boldsymbol{a}=\hat{j}+\hat{k}$ has components $(0,\ 1,\ 1)$ and $\boldsymbol{b}=2\hat{i}+\hat{k}$ has components $(2,\ 0,\ 1)$. Watch the zeros. Substitute into each row of the formula.

$$\begin{pmatrix}0\\1\\1\end{pmatrix}\times\begin{pmatrix}2\\0\\1\end{pmatrix}=\begin{pmatrix}1\times1-1\times0\\1\times2-0\times1\\0\times0-1\times2\end{pmatrix}$$

Simplify each row, then check the answer is perpendicular to both vectors using the tester on the left. This is a technology-active paper, so you can also enter both vectors in your calculator and use its cross product function.

$$=\begin{pmatrix}1\\2\\-2\end{pmatrix}=\hat{i}+2\hat{j}-2\hat{k}$$