Question

The cartesian equations of a line are$6x-2=3y+1=2z-2$. Find its direction ratios and also find a vector equation of the line.

Answer 1

Given that,

$6x-2=3y+1=2z-2$

$6(x-\frac{2}{6})=3(y+\frac{1}{3})=2(z-\frac{2}{2})$

$6(x-\frac{1}{3})=3(y-\left(\frac{-1}{3}\right))=2(z-1)$

$\frac{(x-\frac{1}{3})}{\frac{1}{6}}=\frac{(y-\left(\frac{-1}{3}\right))}{\frac{1}{3}}=\frac{(z-1)}{\frac{1}{2}}$

By comparing with

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

We get

$x_1=\frac{1}{3},y_1=-\frac{1}{3},z_1=1$

$a=\frac{1}{6.},b=\frac{1}{3},c=\frac{1}{2}$

Hence direction ratio are $(a,b,c)=(\frac{1}{6},\frac{1}{3},\frac{1}{2})$

Now, vector equation $=+\lambda$, where $\overrightarrow{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$and $\overrightarrow{b}=a\hat{i}+b\hat{j}+c\hat{k}$

$\overrightarrow{r}=(\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k})+\lambda (\frac{1}{6}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k})$