Question

find the distance between the
parallel planes:

a) 2x +y -2z =0 and 2x+y -2z =5

Answer 1

Is that how the question is written?

Answer 2

In Exercises 37 and 38, find the distance between the
parallel planes.
37. 2x + y - 2z = 0 and 2x + y - 2z = 5

Answer 3

This is the formula you have to use

Answer 4

I guess i need that creepy formula: "In the case where the line £ is in IR2 and its equation has the general form
ax + by = c, the distance d(B, €) from B = (x0, y0) is given by the formula"

Answer 5

They dont give you a graph?

Answer 6

\[d(\beta,\iota) = \frac{ |a x_{0} + b y_{0}+ c| }{\sqrt{a ^{2}+b ^{2}} }\]

Answer 7

no, they don't give anything. But i guess it would be something like:

Answer 8

@texaschic101 @inkyvoyd Could you give me a hand on this one..

Answer 9

@paki @welshfella

Answer 10

In general, the distance d(B, <JP) from the point B = (x0, y0, z0) to the plane whose
general equation is ax + by + cz = d is given by the formula:
\[d(\beta, Plane) = \frac{ |ax _{0}+ by _{0}+ c z _{0} - d |}{ \sqrt{a ^{2}+b ^{2}+c ^{2}} }\]

Answer 11

first you must find a point that lies on one of the planes
for the first equation x = y = z = 0 so there is the point (0,0,0) on that plane

Answer 12

so now you can use your formula

Answer 13

that's all ?

Answer 14

what if use a point of the other plane should i get the same answer ?

Answer 15

x0 y0 and z0 = 0 d = 5

Answer 16

yes I've never come across this formula before but i dont see why not
Its easy to use (0,0,0) though

Answer 17

the numerator will be 5
and plug in a = 2 , b = 1 and c = -2

Answer 18

ok it seems easy...
could you help me with this one:
Find the normal form of the equation of the plane that
passes through P = (O, - 2, 5 ) and is parallel to the
plane with general equation 6x - y + 2z = 3 .

Answer 19

I'd have to look all that stuff up I'm afraid but i haven't got the time at the moment.

Answer 20

ok fine, thanks anyway