Find the Cartesian equation of the line passing through $(3,-2,-5)$ and $(3,-2,6)$ in $3$D.
The equation of the line through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by
$$
\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})
$$
where $\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$ and $\vec{b}-\vec{a}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$
$$
\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}
$$
For the line through the points $(3,-2,-5)$ and $(3,-2,6)$ is
$$\lambda=
\boxed{\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}}
$$
Is it the correct solution and how do I make sense of the final equation of the line through the given points ?
Best Answer
I don't think we should divide by zero.
Observe the first two coordinates of the points.
They satisfy $x=3$ and $y=-2$.
The line is the intersection of $x=3$ and $y=-2$.
The formula is only used when $x_1 \ne x_2$, $y_1 \ne y_2$, and $z_1 \ne z_2$.