Question

Find the equation of a line in
$\left(a\right)$ vector and
$\left(b\right)$ Cartesian form which joins the points whose position vector are $\hat{i}+2\hat{j}-3\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$

Answer 1

Let the given points be $A(1,2,-3)$ and $B(2,1,-1)$

then $\overrightarrow{a}=\hat{i}+2\hat{j}-3\hat{k},\overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k}$

Direction ratios of line $AB$ are proportional to $2-1,1-2,-1+3$

i.e $1,-1,2$

(a) vector equation of the line is:

$\overrightarrow{r}=\overrightarrow{a}+\lambda \left(\hat{i}-\hat{j}+2\hat{k}\right)=\left(\hat{i}+2\hat{j}-3\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+2\hat{k}\right)$

(b) cartesian equation of the line is

$\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+3}{2}$