Find the value of p,so that the lines
$l_{1} : \frac{1 - x}{3} = \frac{7 y - 14}{p} = \frac{z - 3}{2} a n d l_{2} : \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are perpendicular to each other. Also find the equations of a line passing through a point $(3, 2, - 4)$ and parallel to line $l_{1}$ .

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Solution

$Given lines are$
$l_{1} : \frac{1 - x}{3} = \frac{7 y - 14}{p} = \frac{z - 3}{2} a n d l_{2} : \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$

$l_{1} : \frac{- (x - 1)}{3} = \frac{7 (y - 2)}{p} = \frac{z - 3}{2} a n d l_{2} : \frac{- 7 (x - 1)}{3 p} = \frac{y - 5}{1} = \frac{- (z - 6)}{5}$

$l_{1} : \frac{x - 1}{- 3} = \frac{y - 2}{\frac{p}{7}} = \frac{z - 3}{2} a n d l_{2} : \frac{x - 1}{- \frac{3 p}{7}} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$
$As lines$ $l_{1} a n d l_{2}$ $are perpendicular$

$∴ - 3 (\frac{- 3 p}{7}) + \frac{p}{7} (1) + 2 (- 5) = 0$

$\Rightarrow \frac{9 p}{7} + \frac{p}{7} = 10$

$\Rightarrow \frac{10 p}{7} = 10 \Rightarrow p = 7$

Equation of line through(3,2-4) and parallel to $l_{1}$ is $\frac{x - 3}{- 3} = \frac{y - 2}{1} = \frac{z + 4}{2}$ .

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Similar questions

l_{1} : \frac{1 - x}{3} = \frac{7 y - 14}{p} = \frac{z - 3}{2}

l_{2} : \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}

(3, 2, - 4)

l_{1}

(3, 2)

x - 2 y = 2

y = 3

(- 2, 4, - 5)

\frac{x + 3}{3} = \frac{4 - y}{5} = \frac{z + 8}{6}

\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} and \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}

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