1. **Problem statement:** Given vectors $\vec{a} = \begin{pmatrix}2 \\ 4\end{pmatrix}$ and $\vec{b} = \begin{pmatrix}-1 \\ 3\end{pmatrix}$, draw the vectors $\vec{a}$, $\vec{b}$, $\vec{a} + \vec{b}$, $\vec{a} - \vec{b}$, $-3\vec{a}$, and $2\vec{a} - 3\vec{b}$. Show how to construct the sum and difference geometrically. 2. **Formulas and rules:** - Vector addition: $\vec{a} + \vec{b} = \begin{pmatrix}a_1 + b_1 \\ a_2 + b_2\end{pmatrix}$ - Vector subtraction: $\vec{a} - \vec{b} = \begin{pmatrix}a_1 - b_1 \\ a_2 - b_2\end{pmatrix}$ - Scalar multiplication: $k\vec{a} = \begin{pmatrix}k a_1 \\ k a_2\end{pmatrix}$ - Geometrically, vector addition corresponds to placing the tail of $\vec{b}$ at the head of $\vec{a}$ and drawing the resultant vector from the tail of $\vec{a}$ to the head of $\vec{b}$. 3. **Calculate each vector:** - $\vec{a} = \begin{pmatrix}2 \\ 4\end{pmatrix}$ - $\vec{b} = \begin{pmatrix}-1 \\ 3\end{pmatrix}$ - $\vec{a} + \vec{b} = \begin{pmatrix}2 + (-1) \\ 4 + 3\end{pmatrix} = \begin{pmatrix}1 \\ 7\end{pmatrix}$ - $\vec{a} - \vec{b} = \begin{pmatrix}2 - (-1) \\ 4 - 3\end{pmatrix} = \begin{pmatrix}3 \\ 1\end{pmatrix}$ - $-3\vec{a} = -3 \times \begin{pmatrix}2 \\ 4\end{pmatrix} = \begin{pmatrix}-6 \\ -12\end{pmatrix}$ - $2\vec{a} - 3\vec{b} = 2 \times \begin{pmatrix}2 \\ 4\end{pmatrix} - 3 \times \begin{pmatrix}-1 \\ 3\end{pmatrix} = \begin{pmatrix}4 \\ 8\end{pmatrix} - \begin{pmatrix}-3 \\ 9\end{pmatrix} = \begin{pmatrix}4 - (-3) \\ 8 - 9\end{pmatrix} = \begin{pmatrix}7 \\ -1\end{pmatrix}$ 4. **Geometric construction:** - To construct $\vec{a} + \vec{b}$, place the tail of $\vec{b}$ at the head of $\vec{a}$ and draw the vector from the tail of $\vec{a}$ to the head of $\vec{b}$. - To construct $\vec{a} - \vec{b}$, place the tail of $-\vec{b}$ (which is $\vec{b}$ reversed) at the head of $\vec{a}$ and draw the vector from the tail of $\vec{a}$ to the head of $-\vec{b}$. Final vectors: - $\vec{a} = \begin{pmatrix}2 \\ 4\end{pmatrix}$ - $\vec{b} = \begin{pmatrix}-1 \\ 3\end{pmatrix}$ - $\vec{a} + \vec{b} = \begin{pmatrix}1 \\ 7\end{pmatrix}$ - $\vec{a} - \vec{b} = \begin{pmatrix}3 \\ 1\end{pmatrix}$ - $-3\vec{a} = \begin{pmatrix}-6 \\ -12\end{pmatrix}$ - $2\vec{a} - 3\vec{b} = \begin{pmatrix}7 \\ -1\end{pmatrix}$