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The Cartesian Equation of a Line Are 3x + 1 = 6y − 2 = 1 − Z. Find the Fixed Point Through Which It Passes, Its Direction Ratios and Also Its Vector Equation.

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प्रश्न

The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.

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उत्तर

he cartesian equation of the given line is 

3x + 1 = 6y − 2 = 1 − z
It can be re-written as

\[\frac{x + \frac{1}{3}}{\frac{1}{3}} = \frac{y - \frac{1}{3}}{\frac{1}{6}} = \frac{z - 1}{- 1}\]

\[ = \frac{x - \left( - \frac{1}{3} \right)}{2} = \frac{y - \frac{1}{3}}{1} = \frac{z - 1}{- 6}\]

Thus, the given line passes through the point

\[\left( - \frac{1}{3}, \frac{1}{3}, 1 \right)\]  and its direction ratios are proportional to 2, 1,-6. It is parallel to the vector \[\overrightarrow{b} = 2 \hat{i} + \hat{j} - 6 \hat{k} \]

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  \[\vec{r} = \vec{a} + \lambda \vec{b}\]

Vector equation of the required line is

\[\overrightarrow{r} = \left( - \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} + \hat{k} \right) + \lambda \left( 2 \hat{i} + \hat{j} - 6 \hat{k} \right)\]

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

 

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अध्याय 27: Straight Line in Space - Exercise 28.1 [पृष्ठ १०]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
Exercise 28.1 | Q 17 | पृष्ठ १०

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