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प्रश्न
Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\]
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उत्तर
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
\[x \hat{i} + y \hat{j} + z \hat{k} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda \left( - 2 \hat{i} + 14 \hat{j} + 3 \hat{k} \right) [\text { Putting } \overrightarrow{r} = x \hat{i} + y \hat{j} + z k \text{ in } (1)]\]
\[ \Rightarrow x \hat{ i }+ y \hat{j} + z \hat{k} = \left( 1 - 2\lambda \right) \hat{i} + \left( 2 + 14 \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \]
\[\text{ Comparing the coefficients of } \hat{i} , \hat{j} \text{ and } \hat{k} , \text{we get } \]
\[x = 1 - 2\lambda, y = 2 + 14 \lambda, z = 3 + 3\lambda\]
\[ \Rightarrow \frac{x - 1}{- 2} = \lambda, \frac{y - 2}{14} = \lambda, \frac{z - 3}{3} = \lambda\]
\[ \Rightarrow \frac{x - 1}{- 2} = \frac{y - 2}{14} = \frac{z - 3}{3} = \lambda\]
\[\text{ Hence, the cartesian form of (1) is } \]
\[\frac{x - 1}{- 2} = \frac{y - 2}{14} = \frac{z - 3}{3}\]
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