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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.
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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