Question

Find an equation of a plane through the point (-5, 1, -1) which is parallel to the plane 2x - 1y - 1z = -2 in which the coefficient of x is 2.

Answer 1

Have you considered starting at the given point and copying the direction numbers from the given plane?

Answer 2

yes i have, and i dont understand how to get the equation right

Answer 3

the normal to the given plane is <2,-1,-1>

Answer 4

i have that, what i have so far is 2(x+5) -1(y-1) -1(z+1)+12, not sure if i made a math error somewhere or what

Answer 5

any plane parallel to 2x - 1y - 1z = -2 is
2x - 1y -1z = d
we are given that it goes through (-5,1,-1) , so plug these in and solve for d

Answer 6

2(-5) -1(1) - 1(-1) = d
-10 - 1 + 1 = d
-11 + 1 = d
-10 = d

Answer 7

2x -1y -1z = -10 is your answer

Answer 8

also you could it your way. The normal to your given plane is <2,-1,-1>, and any plane that has this normal will be parallel to the given plane.
Now the plane that has this normal *and* contains the point (-5,1,-1) is
2 ( x + 5) -1( y -1) -1 ( z + 1) = 0
comes out to the same answer

Answer 9

in general, a plane that has normal <a,b,c> and contains the point (x1,y1,z1)
is
a (x-x1) + b (y - y1) + c (z - z1) = 0

also this has a nice dot product expression
<a,b,c> . <x-x1, y-y1,z-z1> = 0