Question

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

Answer 1

The Cartesian equation of a line is
$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}......\left(1\right)$
The given line passes through the point $(5,-4,6)$. The position vector of this point is $\overrightarrow{a}=5\hat{i}-4\hat{j}+6\hat{k}$.
Also, the direction ratios of the given line are $3,7$ and $2$.
This means that the line is in the direction of vector $\overrightarrow{b}=3\hat{i}+7\hat{j}+2\hat{k}$.
It is known that the line through position vector $\overrightarrow{a}$ and in the direction of the vector $\overrightarrow{b}$ is given by the equation,
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$, $\lambda \in R$
$\Rightarrow$ $\overrightarrow{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda (3\hat{i}+7\hat{j}+2\hat{k})$
This is the required equation of the given line in vector form.