2Vectors
IA Vectors and Matrices
2.9 Geometry
2.9.1 Lines
Any line through a and parallel to t can be written as
x = a + λt.
By crossing both sides of the equation with t, we have
Theorem. The equation of a straight line through a and parallel to t is
(x − a) × t = 0 or x × t = a × t.
2.9.2 Plane
To define a plane Π, we need a normal
n
to the plane and a fixed point
b
. For
any
x ∈
Π, the vector
x − b
is contained in the plane and is thus normal to
n
,
i.e. (x − b) · n = 0.
Theorem. The equation of a plane through b with normal n is given by
x · n = b · n.
If
n = ˆn
is a unit normal, then
d
=
x · ˆn = b · ˆn
is the perpendicular distance
from the origin to Π.
Alternatively, if a, b, c lie in the plane, then the equation of the plane is
(x − a) · [(b − a) × (c − a)] = 0.
Example.
(i)
Consider the intersection between a line
x × t = a × t
with the plane
x · n = b · n. Cross n on the right with the line equation to obtain
(x · n)t − (t · n)x = (a × t) × n
Eliminate x · n using x · n = b · n
(t · n)x = (b · n)t − (a × t) × n
Provided t · n is non-zero, the point of intersection is
x =
(b · n)t − (a × t) × n
t · n
.
Exercise: what if t · n = 0?
(ii)
Shortest distance between two lines. Let
L
1
be (
x − a
1
)
× t
1
=
0
and
L
2
be (x − a
2
) × t
2
= 0.
The distance of closest approach
s
is along a line perpendicular to both
L
1
and
L
2
, i.e. the line of closest approach is perpendicular to both lines and
thus parallel to
t
1
× t
2
. The distance
s
can then be found by projecting
a
1
− a
2
onto t
1
× t
2
. Thus s =
(a
1
− a
2
) ·
t
1
×t
2
|t
1
×t
2
|
.