Question

Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).

Answer 1

We know that the vector equation of a line passing through the points with position vectors `vec a` and `vec b` is   \[\vec{r} = \vec{a} + \lambda \left( \vec{b} - \vec{a} \right)\] , where  \[\lambda\]  is a scalar.
Here,

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

\[ \vec{b} = 2 \hat{i}  + \hat{j} + \hat{ k} \]

Vector equation of the required line is

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

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

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

Reducing (1) to cartesian form, we get

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

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

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

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

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

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

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

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