# Distance between the given parallel planes:

• May 27th 2010, 01:21 PM
sj9110
Distance between the given parallel planes:
I had the following problem on my test today and I supposedly got the wrong answer but I can't seem to find where I went wrong. Please let me know if you see an error in my work:

Find the distance between the given parallel planes: 4z = 6y - 2x (Plane A)
6z = 2 - 3x + 9y (Plane B)

Let x = y = 0 on plane 6z = 2 - 3x + 9
6z = 2
z = 3
Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

plane A has: a = 2, y = -6, z = 4, d = 0

so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

12 / sqrt(56) and further reduced to 6/sqrt(14)

the computer assignment says this is wrong... any help is greatly appreciated.

• May 27th 2010, 01:30 PM
Plato
Quote:

Originally Posted by sj9110
I had the following problem on my test today and I supposedly got the wrong answer but I can't seem to find where I went wrong. Please let me know if you see an error in my work:

Find the distance between the given parallel planes: 4z = 6y - 2x (Plane A)
6z = 2 - 3x + 9y (Plane B)

Let x = y = 0 on plane 6z = 2 - 3x + 9
6z = 2
z = 3
Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

plane A has: a = 2, y = -6, z = 4, d = 0

so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

12 / sqrt(56) and further reduced to 6/sqrt(14)

the computer assignment says this is wrong... any help is greatly appreciated.

I agree with your answer. But try this form: $\displaystyle \frac{3\sqrt{14}}{7}$
• May 27th 2010, 04:11 PM
sj9110
What's the difference? (how did you arrive at that answer?).. also, it was a test so I only had that one shot, what I really want to know is if I have a legitimate claim that I deserve points for this problem so I can bring it up with my professor..
• May 27th 2010, 08:05 PM
11rdc11
Quote:

Originally Posted by sj9110
What's the difference? (how did you arrive at that answer?).. also, it was a test so I only had that one shot, what I really want to know is if I have a legitimate claim that I deserve points for this problem so I can bring it up with my professor..

rationalize the den.

$\displaystyle \bigg(\frac{6}{\sqrt{14}}\bigg)\bigg(\frac{\sqrt{1 4}}{\sqrt{14}}\bigg) = \frac{6\sqrt{14}}{14} = \frac{3\sqrt{14}}{7}$

My teacher marks the answer wrong unless it is the simplest form, so I do not think he will give you credit.
• May 28th 2010, 02:30 AM
HallsofIvy
Quote:

Originally Posted by sj9110
I had the following problem on my test today and I supposedly got the wrong answer but I can't seem to find where I went wrong. Please let me know if you see an error in my work:

Sorry, you are not going to get, and should not get, partial credit.

Quote:

Find the distance between the given parallel planes: 4z = 6y - 2x (Plane A)
6z = 2 - 3x + 9y (Plane B)

Let x = y = 0 on plane 6z = 2 - 3x + 9
6z = 2
z = 3
Here's an obvious error- if 6z= 2, then z= 1/3, not 3!

Quote:

Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

plane A has: a = 2, y = -6, z = 4, d = 0

so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

12 / sqrt(56) and further reduced to 6/sqrt(14)

the computer assignment says this is wrong... any help is greatly appreciated.