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

Vector form

$3\hat{i} -2\hat{j} -5\hat{k}$ +

$\lambda (11\hat{k})$
Catesian form

$\dfrac{x-3}{0}$ =

$\dfrac{y+2}{0}$ =

$\dfrac{z+5}{11}$

Vector form

$3\hat{i} -2\hat{j} + 6\hat{k}$
Catesian form

$\dfrac{x-3}{0}$ =

$\dfrac{y+2}{0}$ =

$\dfrac{z+5}{\lambda}$

Vector form

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

$\dfrac{x-3}{11}$ =

$\dfrac{y+2}{11}$ =

$\dfrac{z+5}{11}$

Vector form

$3\hat{i} -2\hat{j} + 6\hat{k}$
Catesian form

$\dfrac{x-3}{11}$ =

$\dfrac{y+2}{11}$ =

$\dfrac{z+5}{11}$

Karnataka PU 2017

Solution

The correct answer is Vector form

$3\hat{i} -2\hat{j} -5\hat{k}$ +

$\lambda (11\hat{k})$
Catesian form

$\dfrac{x-3}{0}$ =

$\dfrac{y+2}{0}$ =

$\dfrac{z+5}{11}$

Explanation

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

$z = -5 + 11\lambda$

⇒

$\dfrac{x-3}{0}$ =

$\dfrac{y+2}{0}$ =

$\dfrac{z+5}{11}$ =

$\lambda$

∴

$\dfrac{x-3}{0}$ =

$\dfrac{y+2}{0}$ =

$\dfrac{z+5}{11}$