A plane is given by the equation

$\displaystyle 3x-4y+2z+4=0$

Find the equation of a linelwhich lies in the given plane and has distance 6 away from thexy-plane.

The line should be given in the form

$\displaystyle x=t \wedge y=t \wedge z=t$

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- Aug 14th 2008, 02:36 PMMatteNoobPlane with a line
A plane is given by the equation

$\displaystyle 3x-4y+2z+4=0$

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

$\displaystyle x=t \wedge y=t \wedge z=t$ - Aug 15th 2008, 10:55 AMearboth
Construct a plane E parallel to the xy-plane and in a distance of 6 units form the xy-plane:

$\displaystyle E: z = 6$

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

Substitute z = 6 into the given equation:

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

Therefore

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

Remark: This is my last post for the next 4 weeks. Bye and much success! - Aug 16th 2008, 12:27 PMMatteNoob
Thanks a lot! :)

- Aug 18th 2008, 01:20 AMearboth
Thanks for the thanks - but I've forgotten the second solution.

The plane $\displaystyle E_2: z=-6$ 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:

$\displaystyle

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

$