1. **State the problem:** We need to find a vector equation for the line $l_1$ passing through points $A(2,5,9)$ and $B(6,0,10)$. 2. **Formula for vector equation of a line:** The vector equation of a line passing through point $\mathbf{r}_0$ with direction vector $\mathbf{d}$ is $$\mathbf{r} = \mathbf{r}_0 + t\mathbf{d}$$ where $t$ is a scalar parameter. 3. **Find the direction vector $\mathbf{d}$:** $$\mathbf{d} = \overrightarrow{AB} = (6-2, 0-5, 10-9) = (4, -5, 1)$$ 4. **Write the vector equation for $l_1$:** Using point $A$ as $\mathbf{r}_0$, $$\mathbf{r} = \begin{pmatrix}2 \\ 5 \\ 9\end{pmatrix} + t \begin{pmatrix}4 \\ -5 \\ 1\end{pmatrix}$$ --- 5. **Show that point $A$ lies on both $l_1$ and $l_2$:** - For $l_1$, when $t=0$, $$\mathbf{r} = \begin{pmatrix}2 \\ 5 \\ 9\end{pmatrix}$$ which is point $A$. - For $l_2$, the vector equation is $$\mathbf{r} = \begin{pmatrix}8 \\ 8 \\ 0\end{pmatrix} + \mu \begin{pmatrix}2 \\ 1 \\ -3\end{pmatrix}$$ We want to find $\mu$ such that $$\begin{pmatrix}8 + 2\mu \\ 8 + \mu \\ 0 - 3\mu\end{pmatrix} = \begin{pmatrix}2 \\ 5 \\ 9\end{pmatrix}$$ 6. **Solve for $\mu$ component-wise:** - From $x$-component: $8 + 2\mu = 2 \Rightarrow 2\mu = -6 \Rightarrow \mu = -3$ - From $y$-component: $8 + \mu = 5 \Rightarrow \mu = -3$ - From $z$-component: $0 - 3\mu = 9 \Rightarrow -3\mu = 9 \Rightarrow \mu = -3$ All components give $\mu = -3$, confirming point $A$ lies on $l_2$. **Final answers:** - Vector equation of $l_1$: $$\mathbf{r} = \begin{pmatrix}2 \\ 5 \\ 9\end{pmatrix} + t \begin{pmatrix}4 \\ -5 \\ 1\end{pmatrix}$$ - Point $A$ is the intersection of $l_1$ and $l_2$ because it satisfies both equations at $t=0$ and $\mu = -3$ respectively.