Question

The cartesian equation of the line are $3x-1=6y+2=1-z$. Find the vector equation of line.

Answer 1

Given equations of the line are
$3x-1=6y+2=1-z$
$3(x-\frac{1}{3})=6(y+\frac{2}{6})=-(z-1)$
$3(x-\frac{1}{3})=6(y+\frac{2}{6})=-(z-1)$
$\Rightarrow \frac{(x-\frac{1}{3})}{\frac{1}{3}}=\frac{(y+\frac{1}{3})}{\frac{1}{6}}=\frac{(z-1)}{-1}$ ....... (1)
Now consider the general equation of the line
$\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n}$ ......... (2)
Where $l$, m and n are the direction ratios of the line and the point $(a,b,c)$ lies on the line.
Compare the equation (1), with the general equation (2)
We have, $l=\frac{1}{3},m=\frac{1}{6}$ and $n=-1$
Also, $a=\frac{1}{3},b=-\frac{1}{3}$ and $c=1$
This shows that the given line passes through
$(\frac{1}{3},-\frac{1}{3},1)$
Hence, $\overrightarrow{a}=\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k}$ and $\overrightarrow{b}=\frac{1}{3}\hat{i}+\frac{1}{6}\hat{j}-\hat{k}$
The vector equation is given by,
$r=a+\lambda b$
$r=(\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k})+\lambda (\frac{1}{3}\hat{i}+\frac{1}{6}\hat{j}-\hat{k})$