Find the equation of a line in vector form and cartesian form for passing through (3, -2, -5) and (3, -2, 6)

The correct answer is Vector form $3\hat{i} -2\hat{j} -5\hat{k}$ + $\lambda (11\hat{k})$
Catesian form $\dfrac{x-3}{0}$ = $\dfrac{y+2}{0}$ = $\dfrac{z+5}{11}$

Let $A$ (3, -2, -5) and $B$(3, -2, 6) be two points.


$\overrightarrow{a}$ = $3\hat{i} -2\hat{j} -5\hat{k}$

$\overrightarrow{b}$ = $3\hat{i} -2\hat{j} +6\hat{k}$

∴ Vector equation of line AB is $\overrightarrow{r}$ = $\overrightarrow{a}$ + $\lambda (\overrightarrow{b} - \overrightarrow{a})$

= $3\hat{i} -2\hat{j} -5\hat{k}$ + $\lambda (11\hat{k})$



Cartesian form:

Let $\overrightarrow{r}$ = $x\hat{i} +y\hat{j} +z\hat{k}$ = $3\hat{i} -2\hat{j} + (-5 + 11\lambda)\hat{k}$

Equating the co-efficient of $\hat{i}, \hat{j}$ and $\hat{k}$, we get

$x = 3$, $y = -2$ and $z = -5 + 11\lambda$

⇒ $\dfrac{x-3}{0}$ = $\dfrac{y+2}{0}$ = $\dfrac{z+5}{11}$ = $\lambda$

∴ $\dfrac{x-3}{0}$ = $\dfrac{y+2}{0}$ = $\dfrac{z+5}{11}$