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Find the Cartesian Equation of a Line Passing Through (1, −1, 2) and Parallel to the Line Whose Equations Are X − 3 1 = Y − 1 2 = Z + 1 − 2 Also, Reduce the Equation Obtained in Vector Form. - Mathematics

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प्रश्न

Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are  \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\]  Also, reduce the equation obtained in vector form.

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उत्तर

We know that the cartesian equation of a line passing through a point with position vector `vec a `  and parallel to the vector `vec m ` is  \[\frac{x - x_1}{a} = \frac{y - y_2}{b} = \frac{z - z_3}{c}\] 

Here, 

\[\vec{a} = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k} \]

\[ \vec{m} = a \hat{i} + b \hat{j} + c \hat{k} \]

Here,  \[\vec{a} = \hat{i} - \hat{j} + 2 \hat{k} \text{ and } \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}\] 

Cartesian equation of the required line is 

\[\frac{x - 1}{1} = \frac{y - \left( - 1 \right)}{2} = \frac{z - 2}{- 2}\]

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

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

Here, the line is passing through the point  (1,1,-2)  and its direction ratios are proportional to 1, 2, -2 .

Vector equation of the required line is

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

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अध्याय 28: Straight Line in Space - Exercise 28.1 [पृष्ठ १०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.1 | Q 10 | पृष्ठ १०

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The lines  \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\] 

 


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