Question

Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes $\overrightarrow{r}·\left(\hat{i}-\hat{j}+2\hat{k}\right)=5\mathrm{and}\overrightarrow{r}·\left(3\hat{i}+\hat{j}+2\hat{k}\right)=6$

Answer 1

$$\begin{gathered} \text{Let the direction ratios of the required line be proportional to}a,b,c.\text{As it passes through (1, 2, 3), its equations are} \\ \frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c}...\left(1\right) \\ \text{It is given that (1) is parallel to the planes}\overrightarrow{r}\text{.}\left(\hat{i}-\hat{j}+2\hat{k}\right)\text{= 5 and}\overrightarrow{r}\text{.}\left(3\hat{i}+\hat{j}+2\hat{k}\right)\text{= 6 or}x-y+2z=5\text{and}3x+y+2z=6 \\ \Rightarrow a-b+2c=0...\left(2\right) \\ 3a+b+2z=0...\left(3\right) \\ \text{Solving these two by cross-multiplication method, we get} \\ \frac{a}{-2-2}=\frac{b}{6-2}=\frac{c}{1+3} \\ \Rightarrow \frac{a}{-4}=\frac{b}{4}=\frac{c}{4} \\ \Rightarrow \frac{a}{1}=\frac{b}{-1}=\frac{c}{-1}=\lambda \text{(say)} \\ \Rightarrow a=\lambda ;b=-\lambda ;c=-\lambda \\ \text{Substituting these values in (1), we get} \\ \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{-1},\text{which is the Cartesian form of the required line.} \\ \mathbf{\text{Vector form}} \\ \text{The given line passes through a point whose position vector is}\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}\text{and is parallel to the vector}\overrightarrow{b}\text{=}\hat{i}\text{-}\hat{j}\text{+}\hat{k}\text{. So, its equation in vector form is} \\ \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \\ \Rightarrow \overrightarrow{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}\text{-}\hat{j}\text{+}\hat{k}\right) \end{gathered}$$

Disclaimer: The answer given for this problem in the text book is incorrect. The problem should be same as problem #19 to get the text book answer.