Question

Find the vector equation of a line which is parallel to the vector \[2 \hat{i} - \hat{j} + 3 \hat{k}\]  and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.

Answer 1

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

Here, 

\[\vec{a} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k} \]

\[ \vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k} \]
So, the vector equation of the required line is

\[\vec{r} = \left( 5 \hat{i} - 2 \hat{j} + 4 \hat{k} \right) + \lambda \left( 2 \hat{i} - \hat{j} + 3 \hat{k}  \right) . . . (1) \]

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

Reducing (1) to cartesian form, we get

\[x \hat{i} + y \hat{j} + z \hat{k} = \left( 5 \hat{i} - 2 \hat{j} + 4 \hat{k} \right) + \lambda \left( 2 \hat{i} - \hat{j} + 3 \hat{k} \right) [\text{Putting } \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \text   { in }(1)]\]

\[ \Rightarrow x \hat{i} + y \hat{j} + z \hat{k} = \left( 5 + 2\lambda \right) \hat{i} + \left( - 2 - \lambda \right) \hat{j} + \left( 4 + 3\lambda \right) \hat{k} \]

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

\[x = 5 + 2\lambda, y = - 2 - \lambda, z = 4 + 3\lambda\]

\[ \Rightarrow \frac{x - 5}{2} = \lambda, \frac{y + 2}{- 1} = \lambda, \frac{z - 4}{3} = \lambda\]

\[ \Rightarrow \frac{x - 5}{2} = \frac{y + 2}{- 1} = \frac{z - 4}{3} = \lambda\]

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

\[\frac{x - 5}{2} = \frac{y + 2}{- 1} = \frac{z - 4}{3}\]