Question

Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line  \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\] 

Answer 1

We have \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\] 

It can be re-written as 

\[\frac{x + 2}{- 1} = \frac{y + 3}{7} = \frac{z - 3}{\left( \frac{3}{2} \right)}\]

\[ = \frac{x + 2}{- 2} = \frac{y + 3}{14} = \frac{z - 3}{3}\]

This shows that the given line passes through the point  (-2, -3,3) and its direction ratios are proportional to  -2,14,3

Thus, the parallel vector is  \[\overrightarrow{b} = - 2 \hat{i} + 14 \hat{j} + 3 \hat{k} \]

We know that the vector equation of a line passing through a point with position vector

\[\overrightarrow{a} \]and parallel to the vector \[\overrightarrow{b} \] is  \[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}\]

Here,

\[\overrightarrow{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}  \]

\[ \overrightarrow{b} = - 2 \hat{i} + 14 \hat{j} + 3 \hat{k}\]

Vector equation of the required line is 

\[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda \left( - 2 \hat{i} + 14 \hat{j} + 3 \hat{k}  \right) . . . (1)\]

\[\text{ Here } , \lambda \text { is a parameter } . \]

Reducing (1) to cartesian form, we get 

\[x \hat{i} + y \hat{j} + z \hat{k} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) + \lambda \left( - 2 \hat{i} + 14 \hat{j} + 3 \hat{k} \right) [\text { Putting }  \overrightarrow{r} = x \hat{i}  + y \hat{j} + z k \text{ in } (1)]\]

\[ \Rightarrow x \hat{ i }+ y \hat{j} + z \hat{k} = \left( 1 - 2\lambda \right) \hat{i} + \left( 2 + 14 \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \]

\[\text{ Comparing the coefficients of } \hat{i} , \hat{j} \text{ and } \hat{k} , \text{we get } \]

\[x = 1 - 2\lambda, y = 2 + 14 \lambda, z = 3 + 3\lambda\]

\[ \Rightarrow \frac{x - 1}{- 2} = \lambda, \frac{y - 2}{14} = \lambda, \frac{z - 3}{3} = \lambda\]

\[ \Rightarrow \frac{x - 1}{- 2} = \frac{y - 2}{14} = \frac{z - 3}{3} = \lambda\]

\[\text{ Hence, the cartesian form of (1) is } \]

\[\frac{x - 1}{- 2} = \frac{y - 2}{14} = \frac{z - 3}{3}\]