Answer
$\mathbf{r(t)} =3(1-t)\textbf{i}+4(1-t)\textbf{j}, \quad 0 \leq t \leq 1$
Work Step by Step
Step 1
To write the vector equation, we will use the formula: \[ \mathbf{r} = \mathbf{r}_0 + t (\mathbf{r}_1 - \mathbf{r}_0) \] First, we need to determine the vectors $\mathbf{r}_0$ and $\mathbf{r}_1$.
Step 2
We will let $\mathbf{r}_0 = \langle 3, 4 \rangle$ as this is the vector that leads from $\mathbf{Q}$ to $\mathbf{P}$, and the vector $\mathbf{r}_1$ will be the null vector $\langle 0, 0 \rangle$. \[ \begin{align*} \mathbf{r} &= \mathbf{r}_0 + t (\mathbf{r}_1 - \mathbf{r}_0) \\ &= \langle 3, 4 \rangle + t (\langle 0, 0 \rangle - \langle 3, 4 \rangle) \\ &= \langle 3, 4 \rangle + t \langle 0 - 3, 0 - 4 \rangle \\ &= \langle 3, 4 \rangle + t \langle -3, -4 \rangle \end{align*} \] We will also limit the parameter $t$ to range between $0$ and $1$ because the line segment is located between the vectors $\langle 0, 0 \rangle$ and $\langle 3, 4 \rangle$.
Step 3
The vector equation for the line segment is: \[ \mathbf{r(t)} = \langle 3, 4 \rangle + t \langle -3, -4 \rangle=3(1-t)\textbf{i}+4(1-t)\textbf{j}, \quad 0 \leq t \leq 1 \] Result \[ \mathbf{r(t)} = 3(1-t)\textbf{i}+4(1-t)\textbf{j}, \quad 0 \leq t \leq 1 \]
