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प्रश्न
Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by \[\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} .\]
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उत्तर
We know that the cartesian equation of a line passing through a point with position vector \[\overrightarrow{a}\] and parallel to the vector \[\overrightarrow{b}\] is
\[\frac{x - x_1}{a} = \frac{y - y_2}{b} = \frac{z - z_3}{c}\]
Here,
\[\overrightarrow{a} = - 2 \hat{i} + 4j - 5 \hat{k} \]
\[ \overrightarrow{b} = 3 \hat{i} + 5 \hat{j} - 6 \hat{k} \]
The cartesian equation of the required line is
\[\frac{x - \left( - 2 \right)}{3} = \frac{y - 4}{5} = \frac{z - \left( - 5 \right)}{6}\]
\[ = \frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}\]
\[\]
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