Question

Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j}  - 5 \hat{k}  \right) + 9 = 0 .\]

 

Answer 1

\[ \text{ Let a, b, c be the direction ratios of the given line.} \]

\[ \text{ Since the line passes through the point (1, 2, 3) is } ,\]

\[\frac{x - 1}{a} = \frac{y - 2}{b} = \frac{z - 3}{c} . . . \left( 1 \right)\]

\[\text{ Since this line is perpendicular to the plane } \vec{r} .\left( \hat{i} + 2 \hat{j}  - 5 \hat{k}  \right)+ \text{ 9 = 0 or x + 2y - 5z + 9 = 0, the line is parallel to the normal of the plane } .\]

\[\text{ So, the direction ratios of the line are proportional to the direction ratios of the given plane. } \]

\[\text{ So, } \frac{a}{1} = \frac{b}{2} = \frac{c}{- 5} = \lambda\]

\[ \Rightarrow a = \lambda; b = 2\lambda; c = - 5\lambda\]

\[\text{ Substituting these values in (1), we get } \]

\[\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z - 2}{- 5}, \text{ which is the Cartesian form of the line } .\]

\[\text{ Vector form } \]

\[\text{ The given line passes through a point whose position vector is }  \vec{a} = \hat{i}  + 2 \hat{j}  + 3 \hat{k} \text{ and is parallel to the vector } \vec{b} = \hat{i}  + 2 \hat{j}  - 5 \hat{k}  . \text{ So, its equation in vector form is } \]

\[ r^\to = \vec{a} + \lambda \vec{b} \]

\[ \Rightarrow \vec{r} = \left( \hat{i}  + 2 \hat{j} + 3 \hat{k}  \right) + \lambda\left( \hat{i}  + 2 \hat{j} - 5 \hat{k}  \right)\]