Question

Find the cartesian and vector equations of the line which passes through the point (‒2, 4, ‒5) and parallel to the line given by (x + 3)/3 = (y ‒ 4)/5 = (8 ‒ z)/‒6.

Answer 1

Here it is..
Hope it helps..

Answer 2

Answer:

The he Cartesian and vector form are $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$ $\overrightarrow{r}=-2\hat{i}+4\hat{j}-5\hat{k}+\lambda (3\hat{i}+5\hat{j}+6\hat{k})$ respectively.

Step-by-step explanation:

The line which passes through the point (‒2, 4, ‒5). The position vector is

$\overrightarrow{a}=-2\hat{i}+4\hat{j}-5\hat{k}$

Since the line is parallel to

$\frac{x+3}{3}=\frac{y-4}{5}=\frac{8-z}{-6}$

Rewrite the equation,

$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z-8}{6}$              .... (1)

The general form of the equation is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$              .... (2)

On comparing (1) and (2), we get

$a=3,b=5,c=6$

It means the line is parallel to

$\overrightarrow{b}=3\hat{i}+5\hat{j}+6\hat{k}$

The equation of line passing through a point with position vector $\overrightarrow{a}$ and parallel to the vector $\overrightarrow{b}$ is,

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$

$\overrightarrow{r}=-2\hat{i}+4\hat{j}-5\hat{k}+\lambda (3\hat{i}+5\hat{j}+6\hat{k})$

The Cartesian equation is

$\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$

Therefore the Cartesian and vector form are $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$ $\overrightarrow{r}=-2\hat{i}+4\hat{j}-5\hat{k}+\lambda (3\hat{i}+5\hat{j}+6\hat{k})$ respectively.