Question

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes  \[\vec{r} \cdot \left( \hat{i}  - \hat{j}  + 2 \hat{k}  \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i}  + \hat{j}  + \hat{k}  \right) = 6 .\]

 

Answer 1

\[ \text{ Let the direction ratios of the required line be proportional to a, b, c . It passes through (1, 2, 3). So its equations are } \]

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

\[ \text{ It is given that (1) is parallel to the planes }  \vec{r} .\left( \hat{i} - \hat{j} + 2 \hat{k}  \right)= 5 \text{ and }  \vec{r} .\left( 3 \hat{i}  + \hat{j}  + 2 \hat{k}  \right)= 6\text{  or x - y + 2z = 5 and 3x + y + 2z = 6 } \]

\[\text{ Thus },\]

\[a - b + 2c = 0 . . . \left( 2 \right)\]

\[ 3a + b + z = 0 . . . \left( 3 \right)\]

\[\text{ Solving these two by cross-multiplication method, we get } \]

\[\frac{a}{- 1 - 2} = \frac{b}{6 - 1} = \frac{c}{1 + 3}\]

\[ \Rightarrow \frac{a}{- 3} = \frac{b}{5} = \frac{c}{4} = \lambda(\text{ say} )\]

\[ \Rightarrow a = - 3\lambda; b = 5\lambda; c = 4\lambda\]

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

\[\frac{x - 1}{- 3} = \frac{y - 2}{5} = \frac{z - 3}{4}, \text{ which is the Cartesian form of the required line.} \]