Show that the distance between the parallel planes ax +by+cz=d1 and ax +by+cz=d2 is D= (the length of d1-d...?

Question

Show that the distance between the parallel planes ax +by+cz=d1 and ax +by+cz=d2 is D= (the length of d1-d2)/ sqrt(a^2 +b^2 +c^2)

Answer 1

To find the distance between to parallel planes pick an arbitrary point in one plane and find the distance from that point to the other plane. Since the planes are parallel the distance from all the points is the same.

ax + by + cz - d1 = 0
ax + by + cz - d2 = 0

Pick a point P from the first plane. P(d1/a, 0, 0).

Use the distance formula to calculate the distance from P to the second plane.

Distance = | a(d1/a) + b*0 + c*0 - d2 | / √(a² + b² + c²)

Distance = | d1 - d2 | / √(a² + b² + c²)

Answer 2

you're arranged to do it by using potential of breaking the undertaking down into 2 steps. the 1st step million is to verify the section between the two factors 2 dimensionally - indoors the X and Y axes. attempt this by using potential of: sq. Root [(x2-x1)²+(y2-y1)²(x2-x1)²+(y2-y1)²(x2-x1)²+(y2-y1)²(x2-x1)²+(y2-y1)²] From there you're able to bypass directly to Step 2, this is to verify the section 3 dimensionally: Square Root [(Step a million value)²+(z2-z1)²(Step a million value)²+(z2-z1)²(the 1st step million value)²+(z2-z1)²(the 1st step million value)²+(z2-z1)²]