Question

The equation of a line passing through (1, 2, 3) and normal to the plane $2x-3y+6z=11$ is _______________.

Answer 1

Since, line passes through (1, 2, 3)
i.e. (x₁, y₁, z₁) (1, 2, 3) and normal is to the plane 2x − 3y + 6z = 11
∴ Direction co-ordinates are (2, −3, 6) = (a, b, c)
Since equation of line passing through (x₁, y₁, z₁) and in the direction (a, b, c) is
$$\begin{gathered} \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \\ \therefore \mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{is}\mathrm{given}\mathrm{by} \\ \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z-3}{6} \end{gathered}$$