Question

Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2x − y + 3z − 5 = 0.

 

Answer 1

\[ \text{ Let a, b, c be the direction ratios of the given line } .\]
\[ \text{ Since the line passes through the point (1, -1, 2) is } \]
\[\frac{x - 1}{a} = \frac{y + 1}{b} = \frac{z - 2}{c} . . . \left( 1 \right)\]
\[\text{ Since this line is perpendicular to the plane  2x - y + 3z - 5 = 0, the line is parallel to the normal of the plane }.\]
\[\text{ So, the direction ratios of the line are proportional to the direction ratios of the given plane }.\]
\[So,\frac{a}{2} = \frac{b}{- 1} = \frac{c}{3} = \lambda\]
\[ \Rightarrow a = 2\lambda; b = - \lambda; c = 3\lambda\]
\[\text{ Substituting these values in (1), we get }\]
\[\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{3}, \text{ which is the Cartesian form of the line } .\]
\[\text{ Vector form }\]
\[\text{ The given line passes through a point whose position vector is } \vec{a} = \hat{i} - \hat{j}  + 2 \hat{k}  \text{ and is parallel to the vector }  \vec{b} = 2 \hat{i}  - \hat{j}  + 3 \hat{k} .\]
\[\text{ So, its equation in vector form is} \]
\[ \vec{r} = \vec{a} + \lambda \vec{b} \]
\[ \Rightarrow \vec{r} = \left( \hat{i}  - \hat{j}  + 2 \hat{k}  \right) + \lambda\left( 2 \hat{i}  - \hat{j}  + 3 \hat{k}  \right)\]