Get the equation of the line passing through $(1, 2, 1)$ and parallel to $\frac{2 x - 1}{3} = \frac{1 - y}{3} = \frac{3 z - 2}{5}$ .

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Solution

Consider the equation

$\begin{matrix} \frac{2 x - 1}{3} = \frac{1 - y}{3} = \frac{3 z - 2}{5} \frac{2 (x - \frac{1}{2})}{3} = \frac{- (y - 1)}{3} = \frac{z (z - \frac{2}{3})}{5} \frac{x - \frac{1}{2}}{\frac{3}{2}} = \frac{y - 1}{- 3} = \frac{z - \frac{2}{3}}{\frac{5}{3}} \end{matrix}$

Since, the required line is parallel to the given line,

So, its direction ratios are proportion to
$\frac{3}{2}, - 3, \frac{5}{3}$

Therefore, equation of line passing through $(1, 2, 1)$ and is parallel to given line is

$\begin{matrix} \frac{x - 1}{\frac{3}{2}} = \frac{y - 2}{- 3} = \frac{z - 1}{\frac{5}{3}} \frac{2 (x - 1)}{3} = \frac{2 - y}{3} = \frac{3 (z - 1)}{5} \end{matrix}$

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Similar questions

(1, - 2, - 1)

2 x + y = 3

y + z = 1,

x + y + z + 3 = 0 = 2 x - y + 3 z + 1

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

Equation of line passing through the point (2, 3, 1) and parallel to the line of intersection of the plane x – 2y – z + 5 = 0 and x + y + 3z = 6 is

x - y + 3 z = 4

2 x + y + 3 z = 5

x + y + z = 1

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