Write the direction cosines of the line whose cartesian equations… - Vidyadip

Question

Write the direction cosines of the line whose cartesian equations are 6x -2 = 3y + 1 =2z - 4.

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Answer

We have
6x -2 = 3y + 1 =2z - 4
The equation of given line can be re-written as
$\frac{\text{x}-\frac{1}{3}}{\frac{1}{6}}=\frac{\text{y}+\frac{1}{3}}{\frac{1}{3}}=\frac{\text{z}-2}{\frac{1}{2}}$
$\Rightarrow\frac{\text{x}-\frac{1}{3}}{1}=\frac{\text{y}+\frac{1}{3}}{2}=\frac{\text{z}-2}{3}$
The direction ratios of the line parallel to AB are proportional to 1, 2, 3.
Hence, the direction cosines of the line parallel to AB are proportional to
$\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}}$
$=\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$

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