Question

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

Answer 1

$$\begin{gathered} \text{Let}a,b,c\text{be the direction ratios of the given line.} \\ \text{Since the line passes through the point (1, 2, 3) is,} \\ \frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c}...\left(1\right) \\ \text{Since this line is perpendicular to the plane}\overrightarrow{r}\text{.}\left(\hat{i}+2\hat{j}-5\hat{k}\right)\text{+9 = 0 or}x+2y-5z+9=0,\text{the line is parallel to the normal of the plane.} \\ \text{So, the direction ratios of the line are proportional to the direction ratios of the given plane.} \\ \text{So,}\frac{a}{1}=\frac{b}{2}=\frac{c}{-5}=\lambda \\ \Rightarrow a=\lambda ;b=2\lambda ;c=-5\lambda \\ \text{Substituting these values in (1), we get} \\ \frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-5},\text{which is the Cartesian form of the line.} \\ \mathbf{\text{Vector form}} \\ \text{The given line passes through a point whose position vector is}\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}\text{and is parallel to the vector}\overrightarrow{b}\text{=}\hat{i}\text{+ 2}\hat{j}\text{- 5}\hat{k}\text{. So, its equation in vector form is} \\ \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \\ \Rightarrow \overrightarrow{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}\text{+ 2}\hat{j}\text{- 5}\hat{k}\right) \end{gathered}$$