Question

​The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is ____________.

Answer 1

Equation in cartesian form is

$$\begin{gathered} \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \\ \mathrm{i}.\mathrm{e}.\frac{x-5}{3}=\frac{y-\left(-4\right)}{7}=\frac{z-6}{2} \\ \therefore \mathrm{Equation}\mathrm{passes}\mathrm{through}\left(5,-4,6\right)\mathrm{and}\mathrm{direction}\mathrm{ratios}\mathrm{are}\left(3,7,2\right) \\ \therefore \mathrm{Vector}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{is} \\ \overrightarrow{r}=\left(5\hat{i}-4\hat{j}+6\hat{k}\right)+\lambda \left(3\hat{i}+7\hat{j}+2\hat{k}\right) \end{gathered}$$