Find the distance between the given parallel planes.
4z = 4y − 4x, 6z = 3 − 6x + 6y
pick a point in plane #1, say (2,3,1)
Now just use the distance formula to get the distance to
6x-6y+6z=3
or, equivalently,
2x-2y+2z-1 = 0
d = |2*2-2*3+2*1-1|/√(2^2+2^2+1^2) = 1/√5
To find the distance between the given parallel planes, we can find the normal vector of one of the planes, and then calculate the distance between the planes using the formula:
Distance = |(A * x1 + B * y1 + C * z1 + D)| / sqrt(A^2 + B^2 + C^2),
where (A, B, C) is the normal vector of the plane, (x1, y1, z1) is a point on the plane, and D is the constant term in the equation of the plane.
Let's find the normal vector and a point on the first plane:
4z = 4y - 4x
We can rewrite this equation as:
4x + 4y - 4z = 0
Comparing this equation with the general form of a plane equation Ax + By + Cz + D = 0, we can see that A = 4, B = 4, C = -4, and D = 0.
So, the normal vector of the first plane is (4, 4, -4), and a point on the plane is (0, 0, 0).
Now, let's calculate the distance between the two planes using the formula:
Distance = |(A * x1 + B * y1 + C * z1 + D)| / sqrt(A^2 + B^2 + C^2)
Substituting the values we found:
Distance = |(4 * 0 + 4 * 0 + -4 * 0 + 0)| / sqrt(4^2 + 4^2 + (-4)^2)
= 0 / sqrt(16 + 16 + 16)
= 0 / sqrt(48)
= 0
Therefore, the distance between the given parallel planes is 0.
To find the distance between two parallel planes, we need to first rewrite the given planes in the form Ax + By + Cz + D = 0.
Let's start with the first plane:
4z = 4y − 4x
We can rewrite it as:
4x + 4y - 4z = 0
Now let's rewrite the second plane:
6z = 3 − 6x + 6y
We can rewrite it as:
6x - 6y + 6z - 3 = 0
Now that we have both planes in the desired form, we can identify the coefficients A, B, C for each plane.
For the first plane: A1 = 4, B1 = 4, C1 = -4
For the second plane: A2 = 6, B2 = -6, C2 = 6
Since the planes are parallel, their normal vectors (A, B, C) will be proportional to each other.
Now, we can take the cross product of the normal vectors of the planes to find the direction vector of the line perpendicular to both planes. The direction vector of this line will be perpendicular to both planes and will help us find the distance between them.
Let's calculate the cross product of the normal vectors:
(A1, B1, C1) × (A2, B2, C2) = (4, 4, -4) × (6, -6, 6)
To find the cross product, we can use the formula:
(u1, u2, u3) × (v1, v2, v3) = (u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1)
Calculating the cross product:
= (4(-6) - (-4)6, (-4)(6) - 4(6), 4(6) - 4(-6))
= (-24 + 24, -24 - 24, 24 - (-24))
= (0, -48, 48)
The direction vector of the line perpendicular to both planes is (0, -48, 48).
The distance between the planes is equal to the length of a perpendicular segment drawn between the planes. We can use the formula to find the distance:
distance = |D2 - D1| / sqrt(A^2 + B^2 + C^2)
Where D1 and D2 are the distance from the origin to each plane along their respective normal vectors.
Now, let's find D1 and D2:
D1 = -D of the first plane = -3 in our case
D2 = -D of the second plane = 0 in our case
Plugging the values into the formula:
distance = |0 - (-3)| / sqrt(0^2 + (-48)^2 + 48^2)
= 3 / sqrt(2304 + 2304)
= 3 / sqrt(4608)
= 3 / (68.0)
= 0.0441
Hence, the distance between the given parallel planes is approximately 0.0441 units.