I'm having trouble with this one.
Problem: Find the equations for the planes parallel to x+2y-2z=1 and two units away from it.
If a point (p,q,r) is not on the place Ax+By+Cz+D = 0 then the distance to the plane is
(Ap + Bq + Cr + D)/(sqrt(A^2+B^2+C^2))
So let the new equation be x + 2y - 2z + D = 0
and pick any point which satisfies the given equation, say (1,0,0)
So the distance is (1 + 4 + 4 + C)/sqrt(1+0+0)
= (1+D)/sqrt(5) but this is supposed to be 2
(1+D)/sqrt(5) = 2
D = 2SQRT(5) - 1
and your new equation is
x + 2y - 2z + 2SQRT(5) - 1 = 0
I put the 4's in the wrong place, it should have said
"So the distance is (1 + 0 + 0 + C)/sqrt(1+4+4)
= (1+D)/sqrt(5) but this is supposed to be 2 "
the rest of the post is ok
To find the equations for the planes parallel to the given plane and two units away from it, follow these steps:
Step 1: Write the given equation in general form.
The equation x + 2y - 2z = 1 can be rewritten as:
x + 2y - 2z - 1 = 0
Step 2: Find the normal vector of the given plane.
The coefficients of x, y, and z in the general form of the equation give us the normal vector. In this case, the normal vector is (1, 2, -2).
Step 3: Normalize the normal vector.
Divide the normal vector by its magnitude to obtain a unit vector. To normalize the vector (1, 2, -2), calculate its magnitude:
Magnitude = √(1^2 + 2^2 + (-2)^2) = √9 = 3
Dividing each component of the normal vector by its magnitude, we get the normalized normal vector: (1/3, 2/3, -2/3).
Step 4: Find the equation of the plane parallel to the given plane and two units away.
For a plane parallel to the given plane and displaced by distance d, the equation of the new plane is obtained by adding or subtracting d to the constant term of the general form of the equation.
Since the given plane is two units away, we will subtract 2 from the constant term (which is -1) to get -3.
Therefore, the equation for the plane parallel to x + 2y - 2z = 1 and two units away from it is:
x + 2y - 2z - 3 = 0
This is the equation of the desired plane.