Question

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

Answer 1

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