Question

Derive the equation of the line in space passing through two given points, both in vector and Cartesian form.

Answer 1

Let $a$ and $b$ be the position vectors of two points $A\left(x_1{y}_1{z}_1\right)$ and vectors of two points $B\left(x_2{y}_2{z}_2\right)$, respectively that are lying on the line.
Let $r$ be the position vector of an arbitrary point $P(z,y,z)$ then $P$ is a point on the line if and only if
$\overrightarrow{AP}=\overrightarrow{OP}-\overrightarrow{OA}=\overrightarrow{r}-\overrightarrow{a}$ and
$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{b}-\overrightarrow{a}$ are collinear vectors.
$\therefore P$ is on the line if and only if
$\overrightarrow{AP}=\lambda \overrightarrow{AB}$
$\therefore \overrightarrow{r}-\overrightarrow{a}=\lambda (\overrightarrow{b}-\overrightarrow{a})$
$\therefore \overrightarrow{r}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a}),\lambda ϵR....\left(1\right)$
which is the vector form of equation of line
Cartesian form:
Let $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k},\overrightarrow{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k};\overrightarrow{b}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}$
Substituting these in equation $\left(1\right)$ we get,
$x\hat{i}+y\hat{j}+z\hat{k}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}+\lambda \left[\right(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1\left)\hat{k}\right]$
$x=x_1+\lambda (x_2-x_1);y=y_1+\lambda (y_2-y_1);z=z_1+\lambda (z_2-z_1)$
Eliminating $\lambda$, we get
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$ which is the cartesian form.