Question

Find the Cartesian equation of the line which passes through the point $(2,-4,5)$ and is parallel to the line $\overline{r}=(-3,4,8)+k(3,5,6),kϵR$.

Answer 1

Let the given point be $A(2,-4,5)$

Equation of the line passing through point $A(2,-4,5)$ is

$\frac{x-2}{a}=\frac{y+4}{b}=\frac{z-5}{c}$ ---- $\left(1\right)$

where $a,b,c$ are the direction ratios of this line

Since, this line is parallel to the given line

So, it is parallel to $3\hat{i}+5\hat{j}+6\hat{k}$

Therefore,

$\frac{a}{3}=\frac{b}{5}=\frac{c}{6}=k$ (say)

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

Putting the values of $a,b,c$ in $\left(1\right)$

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

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

This the required equation of the line .