Write the direction consines of the line whose cartesian equation… - Vidyadip

Question

Write the direction consines of the line whose cartesian equations are 2x = 3y = -z.

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Answer

We have

2x = 3y = -z

The equation of the given line can be re-written as

$\frac{\text{x}}{\frac{1}{2}}=\frac{\text{y}}{\frac{1}{3}}=\frac{\text{z}}{-1}$

$\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$

The diraction ratios of the line parallel to AB are proportional to 3, 2, -6.

Hence, the direction cosines of the line parallel to AB are proportional to

$\frac{3}{\sqrt{3^2+2^2+(-6)^2}},\frac{2}{\sqrt{3^2+2^2+(-6)^2}},\frac{-6}{\sqrt{3^2+2^2+(-6)^2}}$

$=\frac{3}{7},\frac{2}{7},-\frac{6}{7}$

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