Find the equation of line which passes through $(1, 2, 3)$ and parallel to the line $\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2 z - 6}{3}$ .

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Solution

Line is $∥$ to given line.
$\Rightarrow$ Direction cosine of line is $(1, 7, 3 / 2)$

Equation of line in cartesian form

$\frac{x - 1}{1} = \frac{y - 2}{7} = \frac{z - 3}{3 / 2}$

$\Rightarrow \frac{x - 1}{1} = \frac{y - 2}{7} = \frac{2 z - 6}{3}$

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Similar questions

\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2 z - 6}{3} .

(1, 2, 3)

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

\frac{x - 1}{3} = \frac{y}{2} = \frac{z}{6}

The equation of the line passing through (1, 2, 3) and parallel to the planes x - y + 2z = 5 and 3x + y + z = 6, is
[DSSE 1986]

\frac{x - 1}{- 1} = \frac{y - 2}{3} = \frac{z - 4}{6}

\frac{x}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}

(1, 2, 3), (0, - 1, 0)

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

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