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प्रश्न
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).
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उत्तर
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
\[x \hat{i} + y \hat{j} + z \hat{k} = \left( \hat{i} + 2 \hat{j} - \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + 2 \hat{k} \right) [\text{ Putting } \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \text{ in } (1)]\]
\[ \Rightarrow x \hat{i} + y \hat{j} + z \hat{k} = \left( 1 + \lambda \right) \hat{i} + \left( 2 - \lambda \right) \hat{j} + \left( - 1 + 2\lambda \right) \hat{k} \]
\[\text{ Comparing the coefficients of } \hat{i} , \hat{j} \text{ and } \hat{k} , \text{ we get }\]
\[x = 1 + \lambda, y = 2 - \lambda, z = - 1 + 2\lambda\]
\[ \Rightarrow x - 1 = \lambda, \frac{y - 2}{- 1} = \lambda, \frac{z + 1}{2} = \lambda\]
\[ \Rightarrow \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{2} = \lambda\]
\[\text{ Hence, the cartesian form of (1) is} \]
\[\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{2}\]
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