Find the value of $p,$ so that the lines $l_{1} : \frac{1 + x}{3} = \frac{7 y - 14}{p} = \frac{z - 3}{2}$ and $l_{2} : \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ 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

$l_{1} = \frac{1 - x}{3} = \frac{7 y - 14}{p} = \frac{z - 3}{2}$
$l_{2} = \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$
Given point $P (3, 2, - 4)$
Direction cosines;
$l_{1} = (3, p, 2)$
$l_{2} = (3 p, 1, 5)$
Given that both of the lines are $⊥ .$
Therefore, $(3, p, 2) . (3 p, 1, 5) = 0$
$9 p + p + 10 = 0$
$10 p = - 10$
$p = - 1$
Therefore direction cosine of line $d c_{1} = (3, - 1, 2)$
Direction cosine of line $l_{2} = (- 3, 1, 5)$
Required equation of line which is passing through $P$ and parallel to $l_{1}$ $= (3, 2, - 4) + λ (3, - 1, 2) .$

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Similar questions

Q. 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}

Q. The lines

\frac{1 - x}{3} = \frac{7 y - 14}{2 P} = \frac{z - 3}{2}

and

\frac{7 - 7 x}{3 P} = \frac{y - 5}{1} = \frac{6 - z}{5}

are perpendicular to each other. Find the value of

P

Find the value of p so that the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles

Q. Find the value of p so the line

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

and

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

are at right angles.

Q. Let

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

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

The equation of the line perpendicular to

L_{1}

and

L_{2}

and passing through the point of intersection of

L_{1}

and

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Find the value of p, so that the lines l1:1+x3=7y−14p=z−32 and l2:7−7x3p=y−51=6−z5 are perpendicular to each other. Also find the equations of a line passing through a point (3,2,−4) and parallel to line l1.