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

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

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

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

$l_{2} : \frac{x - 1}{\frac{- 3 p}{7}} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$
both line are perpendicular
$(- 3) (\frac{- 3 p}{7}) + \frac{p}{7} (1) + 2 (- 5) = 0$

$\frac{9 p}{7} + \frac{p}{7} - 10 = 0$

$\frac{9 p + p - 70}{7} = 0$
$10 p = 70$
$p = 7$
eq of line parallel to $l_{1}$ and passing through point $P (3, 2, - 4)$
$\to r = 3^i + 2^j - 4^k + t (- 3^i +^j + 2^k)$

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Similar questions

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

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}

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

L_{2}

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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