Question

If the cartesian equation of a line are $\frac{3-x}{5}=\frac{y+4}{7}=\frac{2z-6}{4}$, write the vector equation for the line.

Answer 1

We have $\frac{3-x}{5}=\frac{y+4}{7}=\frac{2z-6}{4}\Rightarrow \frac{x-3}{5}=\frac{y-(-4)}{7}=\frac{z-3}{2}$

On comparing to $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ we have the direction ratios as -5, 7, 2, and point on the line as (3, -4, 3).

Therefore the required vector equation of line is $\overrightarrow{r}=3\hat{i}-4\hat{j}+3\hat{k}+\lambda (2\hat{k}-5\hat{i}+7\hat{j})$.