# give an equation for the plane that is parallel to the plane....

• Dec 5th 2010, 04:08 PM
break
give an equation for the plane that is parallel to the plane....
give an equation for the plane that is parallel to the plane 5x-4y+z=1 and that passes through the point (2,-1,-2).

i used the formula

f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b) + fz(a,b)(z-c)

and get

(-5x+4y+1) + (-5)(x-2) + (4)(y+1) + (1)(z+2)

then plug in the points for the rest of the variables

-13 + (-5x+10) + (4y+4) + (z+2)

-5x+4y+z+3

i'm pretty sure this is wrong and i'm not sure if i used the right formula for this. can anyone explain to me how to do this? is there a simpler way? thanks!
• Dec 5th 2010, 04:21 PM
Plato
You are making it far too hard.
\$\displaystyle 5x-4y+z=5(2)-4(-1)+1(-2)\$. DONE!
• Dec 5th 2010, 04:57 PM
SammyS
give an equation for the plane that is parallel to the plane....
Quote:

Originally Posted by break
Give an equation for the plane that is parallel to the plane 5x-4y+z=1 and that passes through the point (2,-1,-2).

i used the formula

f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b) + fz(a,b)(z-c)

and get

(-5x+4y+1) + (-5)(x-2) + (4)(y+1) + (1)(z+2)

then plug in the points for the rest of the variables

-13 + (-5x+10) + (4y+4) + (z+2)

-5x+4y+z+3

I'm pretty sure this is wrong and i'm not sure if i used the right formula for this. Can anyone explain to me how to do this? is there a simpler way? thanks!

Any plane parallel to the plane, \$\displaystyle 5x-4y+z=1\$ will be of the form:

\$\displaystyle 5x-4y+z=D\$,

where \$\displaystyle D\$ is a constant. To find what that constant is, plug in the coordinates of any point in the plane.

That's basically what Plato did in one step.
• Dec 5th 2010, 05:46 PM
break
wow... thanks for clearing this up!!