Question

Find the vector equation of a line passing through the point with position vector  \[\hat{i} - 2 \hat{j} - 3 \hat{k}\]  and parallel to the line joining the points with position vectors  \[\hat{i} - \hat{j} + 4 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 2 \hat{k} .\] Also, find the cartesian equivalent of this equation.

Answer 1

We know that the vector equation of a line passing through a point with position vector `vec a`   and parallel to the vector ` vec b ` is ` \[\vec{r} = \vec{a} + \lambda \vec{b}\] 

Here, 

\[\vec{a} = \hat{i} - 2 \hat{j} - 3 \hat{k} \]

\[ \vec{b} = \left( 2 \hat{i} + \hat{j} + 2 \hat{k}  \right) - \left( \hat{i} - \hat{j} + 4 \hat{k} \right) = \hat{i} + 2 \hat{j} - 2 \hat{k}\] 

Vector equation of the required line is 

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

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

Reducing (1) to cartesian form, we get

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

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

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

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

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

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

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

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