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Let $\pi_1$ and $\pi_2$ be planes in $\mathbb{R^3}$ with resp. equations

$\pi_1: 20x+4y-5z+21=0$

$\pi_2: 20x+4y-5z+105=0$

1.Determine the coordinates of a point that is equidistant from $\pi_1$ and $\pi_2$.

2.Determine an equation for a plane that is equidistant from $\pi_1$ and $\pi_2$.

Does anyone the answer for these, these are the two questions I don't get

Jean Marie
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jackkk
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  • In the equation $\pi2$ divide through by 2, so we have the same $x,y,z$ coefficients for both equations. The midpoint will lie on the plane $20x + 4y- 5z + (\frac {21}{2} + \frac{105}{4}) = 0$ distance $\frac {\frac {105}{4} - \frac{21}{2}}{\sqrt{20^2 + 4^2 + 5^2}}$ – Doug M Aug 03 '16 at 21:09
  • @DougM Im not sure what you mean – jackkk Aug 03 '16 at 21:10
  • was there an edit to the equation to $\pi2$? I could have sworn that it said $40x + 8y - 10z + 105 = 0$! With the new equations of the plane there is a new midpoint and and distance. – Doug M Aug 03 '16 at 21:13
  • @DougM I made no changes to the equations in my edit, merely setting them in MathJax. – John Wayland Bales Aug 03 '16 at 21:16
  • @jackkk does it seem plausible that if you have two planes with the same coefficients for $x,y,z$ but different constant terms, that if you split the constant terms, you will find the plane between them. Then the next bit, how do you find the distance between two planes, you "unitize" the normal vector, and look at the size of the constant terms that correspond with that unitized normal vector. – Doug M Aug 03 '16 at 21:18
  • What @DougM is calling your attention to is the fact that the two planes are parallel and that the plane parallel to both lies half-way between them, so the constant for that plane will be the average of the constants. So find the equation of the half-way plane and pick any point on it and you will have an answer to both questions. – John Wayland Bales Aug 03 '16 at 21:18
  • @JohnWaylandBales Im not sure what you mean, im sorry im trying to teach myself vectors and its really hard – jackkk Aug 03 '16 at 21:36
  • Vectors apply to this problem in the sense that the coefficients of $x,y,z$ are the components of a normal vector to the plane, that is, a vector which is orthogonal to every point of the plane. If normal vectors of two planes are parallel (multiples of each other) then the planes are parallel. So you can see that these two planes are parallel since a normal vector for each is $20,i+4,j-5,k$. The constant terms $21$ and $105$ show that they are separated. The plane half-way between them will have a constant term half-way between $21$ and $105$, namely, $63$. – John Wayland Bales Aug 03 '16 at 21:41
  • @JohnWaylandBales so the answer for 2) is 20x+4y−5z+63=0 right? – jackkk Aug 03 '16 at 21:43
  • Yes, that is correct. Now you can pick any point you wish in that plane for the answer to (1). – John Wayland Bales Aug 03 '16 at 21:43
  • @JohnWaylandBales can you show me how to plug in the point – jackkk Aug 03 '16 at 21:45
  • You can let any two variables equal $0$ and find a value for the third which balances the equation found in (2). On this one that will give a fraction for the third variable, but there is nothing wrong with that. If you want a point with integer coordinates that will take a bit of trial and error. Notice that if you let $y=-2$ you only need values of $x$ and $z$ satisfying $20x-8-5z+63=0$ which reduces to finding values of $20x-5z+55=0$. You should be able to find integer values of $x$ and $z$ to go along with $y=-2$. I just started subtracting multiples of $4$ from 63 until I got to $55$. – John Wayland Bales Aug 03 '16 at 21:54
  • when you find a point on the middle plane, to prove it is equidistant from $\pi_1$ and $\pi_2$ evaluate each of those two equations at the point on the middle plane and divide by the magnitude of the square root of the sums of their coefficients, viz. $\sqrt{20^2+4^2+5^2}$. That will give you the distance from the point to each plane. You should get the same distance in each case. – John Wayland Bales Aug 03 '16 at 22:02
  • http://math.stackexchange.com/questions/554380/how-to-find-the-distance-between-two-planes – John Wayland Bales Aug 03 '16 at 22:09

1 Answers1

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The direction normal to both planes is given by the vector (20,4,-5). Take a point A=(-5,0,1) which is on the second plane and write the parametrization of the straight line that goes through A with direction vector as the normal to the planes: (-5,0,1)+ t(20,4,-5). Now find the intersection point of this line with the first plane: 20(-5+20t)+16t-5(1-5t)+21=0 You get t=84/441. This gives you point B. Now simply find the middle point of the interval connecting both points. The plane we are trying to find is parallel to both planes so it is of the form 20x+4y-5z+d=0 Using the formula for distance between 2 parallel planes, we demand |d1-d|=|d-d2| as d1=21, d2= 105 are independent coefficients (resectively) of the planes so you get 20x+4y-5z+63=0. (You can check the midpoint is on the plane).

breeze
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