Question

find the distance between the two parallel planes 2x-2y + z+3=0 and 4x-4y+2z+5=0

Answer 1

the normal vector \(\vec n\) is <2,-2,3>. so using the unit normal \(\hat n\), pick a random point on each plane, work out the vector \(\vec v\) between those 2 points

then use \(\vec v* \hat n = |\vec v | |\hat n| \cos \theta = |\vec v | \cos \theta\) . ie the dot product is the distance as \( |\vec v | \cos \theta\) is that vector \(\vec v\) projected onto the unit normal.

Answer 2

hey irish can this be it's formula
=(d2-d1)/(\[\sqrt{a^2+b^2+c^2}\])

Answer 3

another method, take the unit normal, and form a line using it as the direction vector anchored to a point on one of the planes ... the z intercept seems convenient.

then using the form x(n) y(n) z(n) we can see where the line intersects with the other plane. the value of n is therefore the distance between them.

Answer 4

anchor: (0,0,-3) from plane 1

unit normal: (2,-2,1)/sqrt(9) -- the z coeff is not 3

x(n) = 0 +2/3 n
y(n) = 0 -2/3 n
z(n) = -3 +1/3 n

plane 2:

4 x(n) -4 y(n) + 2z(n) + 5 = 0

solve for n

Answer 5

\[any~point~ on ~(1)~say~(0,0,-3)\]
perpendicular distance of this point from plane (2) is
\[d=\frac{ 4(0)-4(0)+2(-3)+5 }{ \sqrt{4^2+(-4)^2+2^2} }=\frac{ 5 }{ 6 }\]

Answer 6

correction
\[d=\left| \frac{ 4(0)-4(0)+2(-3)+5 }{ \sqrt{4^2+(-4)^2+2^2} } \right|=\frac{ 1 }{ 6 }\]