Question

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

Answer 1

We have
6x − 2 = 3y + 1 = 2z − 4

The equation of given line can be re-written as

$$\begin{gathered} \frac{x-{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{6}}}=\frac{y+{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{3}}}=\frac{z-2}{{\displaystyle \frac{1}{2}}} \\ \Rightarrow \frac{x-{\displaystyle \frac{1}{3}}}{{\displaystyle 1}}=\frac{y+{\displaystyle \frac{1}{3}}}{2}=\frac{z-2}{3} \end{gathered}$$

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

$$\begin{gathered} \frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}} \\ =\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}} \end{gathered}$$.