Lines and Planes in Space
A line is determined by a point on the line and the direction of . The direction of is conveniently described by a vector, so we let u be a vector parallel to . Let be an arbitrary point on and let a and r be the position vectors of and then there is a number such that
r = a + t u
which is a vector equation of . Each value of the parameter gives the position vector of a point on (see the animation below).
If a = ( ), u = ( ), r = ( ), then the vector equation gives us the following parametric equations :
, ,
A plane in space is determined by a point in the plane and a vector n , called normal vector , that is orthogonal to the plane. Let be an arbitrary point in the plane and let a and r be the position vectors of and then
nĦE ( ra ) = 0 .
If a = ( ), n = ( ), r = ( ), then we get the scalar equation of the plane
By collecting terms, we can rewrite the equation of the plane as
.
Two planes are parallel if their normal vectors are parallel and two planes are orthogonal if their normal vectors are orthogonal.
Let be a point in the given plane and let b be the vector from P to , that is,
b = ( )
From the graph below, we see that the distance from to to the plane is equal to the absolute value of the scalar projection of b onto the normal vector n = ( ).
Thus, the distance equals to | n ĦE b | / || n ||
=
=
=
Moreover, the point on that attains the distance is
() (why ??).
Example
Find the point on the plane that is closest to the point ().
Hence, ( ) is the point on the plane ,
that is closest to the point ().
Example
Find the distance between the parallel planes and .
Note that the planes are parallel because their normal vectors
( ) and ( )
are parallel. To find the distance between the planes, we choose any point on one plane and calculate its distance to the plane. It is easy to see that ( ) is on .
By the distance formula, the distance from ( ) to is
.
So the distance between the planes is .
Example
Find the distance between the lines
:
:
Since the two lines are skew, they can be viewed as lying on two parallel planes and . The distance between and is the same as the distance between and .
The common normal vector to both and must be orthogonal to both
u = ( ) and v=().
So a normal vector is
Put in the equation of , we get a point ( ) on and so the equation for is
Since ( ) is on , so the distance between and is the same as the distance from ( ) to . That is,