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Question
Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\] Reduce the corresponding equation in cartesian from.
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Solution
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,
\[\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
\[ \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k} \]
Vector equation of the required line is
\[\vec{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda \left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right) . . . (1)\]
\[\text{Here }, \lambda \text{ is a parameter } . \]
Reducing (1) to cartesian form, we get
\[x \hat{i} + y \hat{j} + z \hat{k} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda \left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right) [\text{ Putting } \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \text{ in } (1)]\]
\[ \Rightarrow x \hat{i} + y \hat{j} + z \hat{k} = \left( 1 + \lambda \right) \hat{i} + \left( 2 - 2\lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \]
\[\text{ Comparing the coefficients of } \hat{i} , \hat{j} \text{ and } \hat{ k} , \text{ we get} \]
\[x = 1 + \lambda, y = 2 - 2\lambda, z = 3 + 3\lambda\]
\[ \Rightarrow x - 1 = \lambda, \frac{y - 2}{- 2} = \lambda, \frac{z - 3}{3} = \lambda\]
\[ \Rightarrow \frac{x - 1}{1} = \frac{y - 2}{- 2} = \frac{z - 3}{3} = \lambda\]
\[\text{ Hence, the cartesian form of } (1) \hspace{0.167em} \text { is } \]
\[\frac{x - 1}{1} = \frac{y - 2}{- 2} = \frac{z - 3}{3}\]
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