Lines and Planes in Space

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A line L is determined by a point P(x[0],y[0],z[0]) on the line and the direction of L.  The direction of L is conveniently described by a vector, so we let u be a vector parallel to L.  Let P(x,y,z)be an arbitrary point on L and let a and r  be the position vectors of P[0] and P then there is a number t such that

r = a + t u

 

which is a vector equation of L . Each value of the parameter t gives the position vector of a point on L (see the animation below).

[Maple Plot]

  3D IMAGE                           

If a = ( x[0], y[0], z[0] ), u = ( a, b, c ), r = ( x, y, z ), then the vector equation gives us the following parametric equations :

x = x[0]+a*t , y = y[0]+b*t , z = z[0]+c*t

 

 

A plane in space is determined by a point P[0](x[1],x[2],x[3]) in the plane and a vector n , called normal vector , that is orthogonal to the plane. Let P(x,y,z) be an arbitrary point in the plane and let a and r be the position vectors of P[0] and P then

nĦE ( r``-``a  ) = 0 .

 

If a = ( x[0], y[0], z[0] ), n = ( a, b, c ), r = ( x, y, z ), then we get the scalar equation of the plane

a*(x-x[0])+b*(y-y[0])+c*(z-z[0]) = 0

 

By collecting terms, we can rewrite the equation of the plane as

  a*x+b*y+c*z+d = 0 .

 

Two planes are parallel if their normal vectors are parallel and two planes are orthogonal if their normal vectors are orthogonal.

 

 

Let P(x[0],y[0],z[0]) be a point in the given plane a*x+b*y+c*z+d = 0 and let b be the vector from P to Q(x[1],y[1],z[1]) , that is,

b = ( x[1]-x[0], y[1]-y[0], z[1]-z[0] )

 

From the graph below, we see that the distance from P to to the plane is equal to the absolute value of the scalar projection of b onto the normal vector n = ( a, b, c ).

[Maple Plot]

3D IMAGE

 

Thus, the distance D equals to  | n ĦE b | / || n ||

 

= abs(a*(x[1]-x[0])+b*(y[1]-y[0])+c*(z[1]-z[0]))/sqrt...

= abs(a*x[1]+b*y[1]+c*z[1]-(a*x[0]+b*y[0]+c*z[0]))/sq...

= abs(a*x[1]+b*y[1]+c*z[1]+d)/sqrt(a^2+b^2+c^2)

 

Moreover, the point on that attains the distance D is

(x[1]+a*D/sqrt(a^2+b^2+c^2), x[2]+b*D/sqrt(a^2+b^2+c...)    (why ??).

 

Example

Find the point on the plane x-2*y-2*z-1 = 0 that is closest to the point (2, 3, 4).

[31/9, 10/9, 1/9]

Hence, ( 31/9, 10/9, 1/9 ) is the point on the plane x-2*y-2*z-1 = 0 ,

that is closest to the point (2, 3, 4).

 

Example

Find the distance between the parallel planes 10*x+2*y-2*z = 10 and 5*x+y-z = 1 .

 

Note that the planes are parallel because their normal vectors

( 10, 2, -2) and ( 5, 1, -1 )

 are parallel. To find the distance D between the planes, we choose any point on one plane and calculate its distance to the plane. It is easy to see that ( 1, 0, 0 ) is on 10*x+2*y-2*z = 5 .

By the distance formula, the distance from ( 1, 0, 0 ) to 5*x+y-z = 1 is

 

abs(5+0-0-1)/sqrt(5^2+1^2+(-1)^2) = 4/(3*sqrt(3)) .

 

So the distance between the planes is 4*sqrt(3)/9 .

[Maple Plot]

3D IMAGE

 

Example

Find the distance between the lines

L[1] : x = 2+t, y = -3+3*t, z = 3-t

L[2] : x = 2*s, y = 1+s, z = -3+4*s

 

Since the two lines are skew, they can be viewed as lying on two parallel planes P[1] and P[2]. The distance between L[1] and L[2] is the same as the distance between P[1] and P[2] .

[Maple Plot]

3D IMAGE

 

The common normal vector to both P[1] and P[2] must be orthogonal to both

u = ( 1, 3, -1 ) and v=(2, 1, 4).  

So a normal vector is

n := vector([13, -6, -5])

 

Put s = 0 in the equation of L[2] , we get a point ( 0, 1, -3 ) on L[2] and so the equation for P[2] is

  13*x-6*y-5*z-9 = 0

 

Since ( 2, -3, 3 ) is on L[1] , so the distance between L[1] and L[2] is the same as the distance from ( 2, -3, 5 ) to 13*x-6*y-5*z+3 = 0. That is,

abs(13*2-6*(-3)-5*3-9)/sqrt(13^2+(-6)^2+(-5)^2) = 2...


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