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

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