Shortest Distance from a Point to a Plane
The shortest distance between a point and a plane is always the length of the line that comes out of the plane at a right angle and meets the point in question. If you are struggling to visualise this try using a piece of paper as the plane and thinking about the shortest distance to points not on that piece of paper.
Therefore the distance we are trying to work out is parallel to the normal vector. If we work out a unit vector in the direction of the normal and calculate the dot product between this unit vector and a line from the pointq to a point on the plane, this will give us the distance we want (The understanding of this is tough and not the important part at this level, but make sure you remember the formulae!).
Consider the plane n.p = d and the point q not on the plane, the unit vector for the vector n is equal to |n|n . So the magnitude of the distance we want, D, is equal to:
D=
n
|n|. ~qp
= |n. ~qp|
|n|
~
qpis the line from the pointq to a point on the plane,p. If we letp= (x0,y0,z0) andq= (x1,y1,z1) this simplifies further to:
D=|(x1−x0, y1−y0, z1−z0).n|
|n|
D=|(x1−x0, y1−y0, z1−z0).n|
|n|
Ifn= (A, B, C) our equation above becomes:
D=
A(x1−x0) +B(y1−y0) +C(z1−z0)
√
A2+B2+C2
As n.p= dwe can rearrange this using that p = (x0, y0, z0) and n = (A, B, C) to give d=−Ax0− By0−Cz0. Therefore:
D=
Ax1+By1+Cz1+d
√
A2+B2+C2
Example:
Find the shortest distance between the plane Π = (1,2,3) +λ(2,5,7) +µ(2,3,6) and the pointq= (4,5,6)Solution: We know the equation for the distance we want is:
D=
Ax1+By1+Cz1+d
√
A2+B2+C2
To find A, B, C and D we need to find the normal vector. To do this we need to do the cross product of the two direction vectors:
n=
i j k 2 5 7 2 3 6
= (9,2,−4)
To find D we us thatn.p=dand therefore:
(9,2,−4).(1,2,3) =d= 1 Therefore the distance we want is equal to:
9(4) + 2(5) +−4(6) + 1
Shortest Distance between Two Skew Lines
When two lines are not parallel and do not intersect we call them skew lines. To find the distance between two skew lines there are various techniques but we will do it by constructing two parallel planes from these skew lines and using this to find the perpendicular distance between them.
If we have two linesL1=A+λ ~B andL2=A0+λ ~B0, to find a normal vector to both of these we just need find the cross product between the direction vector of L1 andL2, B~ ×B~0. After this we can use the plane equationn.p=dwith the normal vector we worked out and a point on each line to construct two parallel planes, Π1and Π2.
To work out the distance between these planes, which is equal to the distance between the original skew lines, we first note that Π1 can be described by n·p = d and Π2 can be described by the equation n·p=d0 and therefore the distance between them is:
|d−d0|
|n|
Example:
Find the shortest distance between the lines L1 : x−2 = 1−y = z+23 and L2 :x+17
3 =−y+ 3 = z+12 .
Click here for a video example
Solution: We can write the equations of the lines in the form:
L1= (2,1,−2) +t(1,−1,3) L2= (−17,3,−1) +s(3,−1,2) The cross product of the direction vectors is:
n=
i j k
1 −1 3 3 −1 2
= (1,7,2)
Constructing these into planes:
Finding Π1:
(1,7,2)·(2,1,−2) = 5 Therefore
Π1: (1,7,2)·p= 5 Finding Π2:
(1,7,2)·(−17,3,−1) = 2 Therefore
Π1: (1,7,2)·q= 2 Therefore the distance we want is:
|5−2|
√12+ 72+ 22 = 3
√54 =
√6 6
Intersection between a Line and a Plane
If we wish to find the point of intersection, p, between a line and a plane then we must obtain the equations for the line and the plane, and then solve as them as we would for simultaneous equations to find a point which lies on both structures.
Example:
Find the point of intersectionP between the line r= (2,3,1) +t(3,3,2) and the planer·(1,−2,4) = 5
Solution: Since we already have the equations for the line and plane, we can proceed to solve as simultaneous equations, by substituting the equation of a point on the line into the equation for the plane.
[(2,3,1) +t(3,3,2)]·(1,−2,4) = 5 (2 + 3t,3 + 3t,1 + 2t)·(1,−2,4) (2 + 3t)−2(3 + 3t) + 4(1 + 2t) = 5
2−6 + 4 + 3t−6t+ 8t= 5
⇒5t= 5
⇒t= 1
This value of the parametert corresponds to the point on the line which intersect with the plane.
Now we simply substitute thistvalue into the equation for the line.
p= (2,3,1) + (3,3,2) = (5,6,2)
Example:
Given the plane2x+ 3y+ 3z= 6
and a line which is perpendicular to the plane and passes through the point Q: (5,3,7), find the point where the line intersects the plane,P.
Solution: From the general equation of a plane r·n=d, we can deduce that a normal vector to the plane is (2,3,3). Since the line is perpendicular to the plane and passes through the pointQ, we can write the vector equation of the line as
r= (5,3,7) +t(2,3,3)
Now, as before, we can solve these equations simultaneously to find the parameter t and hence findP.
2(5 + 2t) + 3(3 + 3t) + 3(7 + 3t) = 6 10 + 4t+ 9 + 9t+ 21 + 9t= 6 10 + 9 + 21 + 4t+ 9t+ 9t= 6
⇒22t=−34
⇒t=34 22 = 17
11 Hence the point of intersection is
89 84 128
Intersection between Two Planes
Two planes, Π1and Π2, which are not parallel to each other will intersect along a straight linel. In order to find the direction vector of l, we can calculate the cross product of the normals of the two planes, giving us a vector parallel to both planes, and hence the direction of the line of intersection. We can then find a point of intersection, i.e. a point on the line, by setting y = 0 and solving the equations of the two planes simultaneously to find xand z values. This will give us our equation for the line of intersection of the two planes.
Example:
Find an equation for the line of intersection of the two planes Π1:x−y−z= 1and
Π2: 2x+ 4y−5z= 5
Click here for a video example
Solution: By considering the standard equation for a plane, r·n = d, we can deduce that the normal vector to Π1 is
n1= (1,−1,−1) and the normal vector to Π2is
n2= (2,4,−5)
Now we take the cross product of these normal vectors to find the direction vector for the line of intersection
n1×n2=
i j k
1 −1 −1
2 4 −5
= (5 + 4,−2 + 5,4 + 2)
= (9,3,6)
So a direction vector of the line of intersection is (3,1,2). Now we find a point which lies on both planes. We sety= 0, and by looking at the equations of the planes, we get the two equations
x−z= 1 and 2x−5z= 5
Types of Intersection between Three Planes
If we have three planes, there are four possibilities:
The three planes intersect each other at a single point.
The three planes intersect along a single straight line.
The planes do not all share any common points but they all intersect.
The planes do not intersect and are parallel to each other.