Answer
$ \mathbf{r(t)} = 2t\mathbf{i}+3t\mathbf{j}+4(1-t)\mathbf{k}, \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 $\mathbf{r}_0 = \langle 0, 0, 4 \rangle$ because this is the vector that points to the point $\mathbf{P} = (0, 0, 4)$, and the vector $\mathbf{r}_1 = \langle 2, 3, 0 \rangle$ will be the vector that points to the point $\mathbf{Q} = (2, 3, 0)$. \[ \begin{align*} \mathbf{r} &= \mathbf{r}_0 + t (\mathbf{r}_1 - \mathbf{r}_0) \\ &= \langle 0, 0, 4 \rangle + t (\langle 2, 3, 0 \rangle - \langle 0, 0, 4 \rangle) \\ &= \langle 0, 0, 4 \rangle + t \langle 2 - 0, 3 - 0, 0 - 4 \rangle \\ &= \langle 0, 0, 4 \rangle + t \langle 2, 3, -4 \rangle \end{align*} \] We will also limit the parameter $t$ to range between $0$ and $1$ because the line segment is located between the vectors $\langle 0, 0, 4 \rangle$ and $\langle 2, 3, 0 \rangle$.
Step 3
The vector equation for the line segment is: \[ \mathbf{r} = \langle 0, 0, 4 \rangle + t \langle 2, 3, -4 \rangle=2t\mathbf{i}+3t\mathbf{j}+4(1-t)\mathbf{k}, \quad 0 \leq t \leq 1 \] Result \[ \mathbf{r} = 2t\mathbf{i}+3t\mathbf{j}+4(1-t)\mathbf{k}, \quad 0 \leq t \leq 1 \]
