Question

The cartasian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$. Find its vector equation.

Answer 1

The given Cartesian equation of the line is

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$

That is,

$\frac{x-5}{3}=\frac{y-\left(-4\right)}{7}=\frac{z-6}{2}$

Therefore,

the line passes through the point whose position vector

$\overrightarrow{a}=5\hat{i}-4\hat{j}+6\hat{k}$

And, parallel to the vector $\overrightarrow{b}=3\hat{i}+7\hat{j}+2\hat{k}$

Thus, the equation is

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$

Therefore,

$\overrightarrow{r}=5\hat{i}-4\hat{j}+6\hat{k}+\lambda \left(3\hat{i}+7\hat{j}+2\hat{k}\right)$

where $\lambda$ is a parametre.