# Find the vector and the cartesian equations of the line that passes through the points $(3, -2, -5), (3, -2, 6).$

$\begin{array}{1 1}\large\frac{x-3}{3}=\large\frac{y+2}{-2}=\large\frac{z+5}{6} \\ \large\frac{x-3}{3}=\large\frac{y+2}{-2}=\large\frac{z+5}{-5} \\ \large\frac{x-3}{0}=\large\frac{y+2}{0}=\large\frac{z+5}{11} \\\large\frac{x-3}{0}=\large\frac{y+2}{0}=\large\frac{z+5}{1} \end{array}$

Toolbox:
• Vector equation of a given line passing through two given point is $\overrightarrow r = \overrightarrow a+t (\overrightarrow b-\overrightarrow a)$ where $t \in R$
• Cartesian equation is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}=\large\frac{z-z_1}{z_2-z_1}$
Let the point $(x_1,y_1,z_1)=(3,-2,-5)$ and $(x_2,y_2,z_2)=(3,-2,6)$
$\overrightarrow b-\overrightarrow a = (3-3)\hat i +(-2+2) \hat j+(6+5) \hat k$
$=11 \hat k$
On substituting for $\overrightarrow a$ and $\overrightarrow b,-\overrightarrow a$ we get,
Vector equation of the line is
$\overrightarrow r = (3\hat i -2 \hat j-5 \hat k)+t(11 \hat k)$
The Cartesian equation is $\large\frac{x-3}{3-3}=\large\frac{y+2}{-2+2}=\large\frac{z+5}{6+5}$
On simplifying we get,
$\large\frac{x-3}{0}=\large\frac{y+2}{0}=\large\frac{z+5}{11}$