Lines
To determine a line in R3, it is sufficient to have a point and a vector parallel to the line.
Suppose we have a fixed point, Po, (xo , yo , zo) and an arbitrary point, P, (x, y, z), on the
line, with a vector, v, parallel to the line, <a, b, c>.
Consider the vectors r, the position vector to the arbitrary point, and ro, the position
vector to the fixed point. If we construct a vector from Po to P, it will be some scalar
times v. It follows that < x - xo , y - yo , z - zo > = tv. This leads to the equations:
Vector
x, y, z  xo  ta, yo  tb, zo  tc
Parametric
x  xo  ta, y  yo  tb, z  zo  tc
Symmetric
x  xo y  yo z  zo


a
b
c
Note that x, y, z are just generic variables, the subscripted x, y, z are numbers for a
specific point on the line. We are using t as a parameter, and, a, b, c are the values from
the vector, v, which is parallel to the line.
Planes
A point and an orthogonal vector determine a plane.
n   r  ro   0
Scalar
a  x  x0   b  y  y0   c  z  z0   0
Linear
ax  by  cz  d  0
As in the line, r is a generic position vector to an arbitrary point on the plane (x, y, z), the
subscripted r is a position vector to a specific point on the plane. The vector, n, is
orthogonal to the plane and is where a, b and c come from.
Problems
1. Angle between planes.
 n n 
  cos 1  1 2 
 n1 n2 
Plane 1 is 5x –3y + 2z = 2, plane 2 is 3x + y – 8z = 5. n1 = <5, -3, 2>, n2 = <3, 1, -8>
n1  n2  4; n1  38; n2  74
4 
  1.6463 radians
 38 74 

  cos 1 
2. Distance between point and plane.
D is the scalar projection of the vector from the point to a point on the plane on to the
normal vector to the plane.
nv
D  compn v 
n
Find the distance from the point, (1, 2, 2) to the plane 3x + 2y – z = 4. Pick any
convenient point on the plane, (1, 1, 1) for instance. The vector v is < 1 – 1, 1 – 2, 1 – 2>
= <0 , -1, -1>. The vector, n (the normal vector to the plane) is given by the coefficients,
1
n = <3, 2, -1>. D 
 0.2673
14
3. Line of intersection for two planes.
Find some point on the line. This line is orthogonal to the normals of both planes, so take
the cross to get a vector parallel to the direction of the line. If the point on the line is (xo,
yo, zo) and the cross of the normals is < a, b, c > then the equation is:
x  x0 y  y0 z  z0


a
b
c
Find the line of intersection for, plane 1, 5x –3y + 2z = 2 and plane 2, 3x + y – 8z = 5.
The normals are n1 = <5, -3, 2>, n2 = <3, 1, -8> The cross product is <22, 46, 14>. To get
a point common to both planes, let z = 0 then solve the system of equations for x and y, x
= 17/14, y = 19/14. The line we want is given by:
x = 17/14 + 22t, y = 19/14 + 46t, z = 14t We can verify this by picking an arbitrary value
for t and substituting into both planes. Let t = 2/7, then x = 7.5, y = 14.5 z = 4.
Plane 1  5(7.5) – 3(14.5) + 2(4) = 2
Plane 2  3(7.5) + 1(14.5) – 8(4) = 5
4. Distance between skew lines.
L1 : x  a  k1t y  b  k2t z  c  k3t
L2 : x  d  k4 s y  e  k5 s z  f  k6 s
The vectors formed by the k’s are parallel to the planes these lines lie on, so their cross
product generates a normal. Picking a value for s, we get a point on L2 which we can use
to get plane 2. Next pick a value for t to get a point on L1, then use the procedure for
point to plane.
L1 : x  1  t y  1  2t z  1  3t
L2 : x  1  s y  2  s z  3  3s
1, 2,3  1,1,3  3, 0, 1  n
Let s  1   2,3, 6  is on the plane.
The plane that contains L2 parallel to the plane containing L1 is given by,
3(x – 3) – (z – 6) = 0. Now let t = 1, this gives the point (2, 3, 4) on the first line, also on
the plane containing that line and parallel to the recently determined plane.
The vector, v, is <(2 – 2, 3 – 3, 6 – 4> = <0, 0, 2>.
nv
2
D  compn v 

 0.6325
n
10