The equation of the line passing through (-4,3,1), parallel to the plane x+2y-z-5=0 and in…

Question

The equation of the line passing through $(-4,3,1)$, parallel to the plane $x+2y-z-5=0$ and intersecting the line $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z-2}{-1}$ is:

Accepted Answer

The line L is parallel to the plane P and intersect with line 4 at point R.

Let the coordinate of point R is, (x1, y1, z1) and it passes through L2.

$${{{x_1} + 1} \over { - 3}} = {{{y_1} - 3} \over 2} = {{{z_1} - 2} \over { - 1}} = t$$

$$ \therefore $$   x1 = -1 - 3t, y1 = 3 + 2t, z1 = 2 - t

$$\overrightarrow {AR}  = \left( {3 - 3t} \right)\widehat i + \left( {2t} \right)\widehat j + \left( {1 - t} \right)\widehat k$$

$$\overrightarrow n  = \widehat i + 2\widehat j - \widehat k$$

As $$\overrightarrow {AR} $$ and $$\overrightarrow n $$ are perpendicular  to each other, So

$$\overrightarrow {AR} $$  $$ \cdot $$  $$\overrightarrow n $$ = 0

$$ \Rightarrow $$  (3 - 3t) 1 + (2t)2 + (1 - t) (- 1) = 0

$$ \Rightarrow $$  3 - 3t + 4t - 1 + t = 0

$$ \Rightarrow $$  2 + 2t = 0

$$ \Rightarrow $$  t = - 1

$$ \therefore $$   point R = (2, 1, 3)

$$ \therefore $$  DR of line L is

= (2 - (- 4), $1$ - 3, 3 - 1)

= (6, - 2, 2)

$$ \therefore $$  Equation of line is

$${{x + 4} \over 6} = {{y - 3} \over { - 2}} = {{z - 1} \over 2}$$

or $${{x + 4} \over 3} = {{y - 3} \over { - 1}} = {{z - 1} \over 1}$$

Solutions · 1