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