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