# Plane with a line

• August 14th 2008, 03:36 PM
MatteNoob
Plane with a line
A plane is given by the equation
$3x-4y+2z+4=0$

Find the equation of a line l which lies in the given plane and has distance 6 away from the xy-plane.

The line should be given in the form
$x=t \wedge y=t \wedge z=t$
• August 15th 2008, 11:55 AM
earboth
Quote:

Originally Posted by MatteNoob
A plane is given by the equation
$P: 3x-4y+2z+4=0$

Find the equation of a line l which lies in the given plane and has distance 6 away from the xy-plane.

The line should be given in the form
$x=t \wedge y=t \wedge z=t$

Construct a plane E parallel to the xy-plane and in a distance of 6 units form the xy-plane:

$E: z = 6$

The line you are looking for is $l = P \cap E$ :

Substitute z = 6 into the given equation:

$3x-4y+2\cdot 6 + 4 = 0~\implies~ y = \frac34x + 4$

Therefore

$l:\left\{ \begin {array}{l}x = t \\ y=\frac34t+4 \\ z=6 \end{array}\right.$

Remark: This is my last post for the next 4 weeks. Bye and much success!
• August 16th 2008, 01:27 PM
MatteNoob
Thanks a lot! :)
• August 18th 2008, 02:20 AM
earboth
Quote:

Originally Posted by MatteNoob
Thanks a lot! :)

Thanks for the thanks - but I've forgotten the second solution.

The plane $E_2: z=-6$ is parallel to the xy-plane too and has also the distance of 6 units to the xy-plane.

Therefore you'll get a second line which satisfies the given conditions:

$
l_2:\left\{ \begin {array}{l}x = t \\ y=\frac34t-2\\ z=-6 \end{array}\right.
$