The equation of a line is $5 x - 3 = 15 y + 7 = 3 - 10 z$ . Write the direction cosines of the line.

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Solution

$5 x - 3 = 15 y + 7 = 3 - 10 z \Rightarrow \frac{x - \frac{3}{5}}{\frac{1}{5}} = \frac{y - (- \frac{7}{15})}{\frac{1}{15}} = \frac{z - \frac{3}{10}}{- \frac{1}{10}}$ .
Comparing this with the standard equation of straight line:
$\frac{x - a_{1}}{b_{1}} = \frac{y - a_{2}}{b_{2}} = \frac{z - a_{3}}{b_{3}}$ , we get, $b_{1} = \frac{1}{5} {, b}_{2} = \frac{1}{15} {, b}_{3} = - \frac{1}{10}$ .
The direction cosines are the components of the unit vector
$^b = \frac{b_{1}^i + b_{2}^j + b_{3}^k}{\sqrt{b_{1}^{2} + b_{2}^{2} + b_{3}^{2}}} = \frac{\frac{1}{5}^i + \frac{1}{15}^j - \frac{1}{10}^k}{\frac{7}{30}} = \frac{6}{7}^i + \frac{2}{7}^j - \frac{3}{7}^k$ .
So, the direction cosines are $\frac{6}{7}, \frac{2}{7}, \frac{- 3}{7} .$

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