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The scalar triple product gives the volume of the parallelopiped whose sides are represented by the vectors a, b, and c. a b c β cosβ c.
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Revision of vector algebra, scalar product, vector product
Triple products, multiple products, applications to geometry
Differentiation of vector functions, applications to mechanics
Scalar and vector fields. Line, surface and volume integrals, curvilinear co-ordinates
Vector operators — grad, div and curl
Vector Identities, curvilinear co-ordinate systems
Gauss’ and Stokes’ Theorems and extensions
Engineering Applications
In which we discuss ...
Vector products:Scalar Triple Product, Vector Triple Product, Vector Quadruple Product
Geometry of Lines and Planes
Solving vector equations
Angular velocity and moments
Scalar triple product given by the true determinant
a
b
c
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
Your knowledge of determinants tells you that if you
swap one pair of rows of a determinant, sign changes;
swap two pairs of rows, its sign stays the same.
Hence
(i)
a
b
c
c
a
b
b
c
a
) (Cyclic permutation.)
(ii)
a
b
c
b
a
c
) and so on. (Anti-cyclic permutation)
(iii) The fact that
a
b
c
a
b
c
allows the scalar triple product to be written as [
a
b
c
This notation is not very helpful, and we will try to avoid it below.
The scalar triple product gives the volume of the parallelopiped whose sides are represented by the vectors
a
b
, and
c
a
b
c
cos
β
c
Vector product (
a
b
) has
magnitude equal to the area of the basedirection perpendicular to the base.
The
component
of
c
in this direction is equal to the height of the parallelopiped
Hence
a
b
c
= volume of parallelopied
a
b
c
) is perpendicular to (
b
c
but (
b
c
) is perpendicular to
b
and
c
So
a
b
c
) must be
coplanar
with
b
and
c
a
b
c
λ
b
μ
c
x
a In arbitrary direction
c
b
a
b
c
( b
x
x
c )
a
b
c
1
a
2
b
c
3
a
3
b
c
2
a
2
b
1
c
2
b
2
c
1
a
3
b
1
c
3
b
3
c
1
a
2
c
2
a
3
c
3
b
1
a
2
b
2
a
3
b
3
c
1
a
1
c
1
a
2
c
2
a
3
c
3
b
1
a
1
b
1
a
2
b
2
a
3
b
3
c
1
a
c
b
1
a
b
c
1
Similarly for components 2 and 3: so
a
b
c
a
c
b
a
b
c
Books say that the vector projection of any old vector
v
into a plane with
normal
ˆn
is
v
INPLANE
ˆn
v
ˆn
The component of
v
in the
ˆn
direction is
v
ˆn
so I would write the vector projection as
v
INPLANE
v
v
ˆn
ˆn
Can we reconcile the two expressions? Subst.
n
a
v
b
ˆn
c
, into our earlier formula
a
b
c
a
c
b
a
b
c
ˆn
v
ˆn
ˆn
ˆn
v
ˆn
v
ˆn
v
v
ˆn
ˆn
Fantastico! But
v
v
n
ˆn
is much easier to understand, cheaper to compute!
Using just the R-H sides of what we just wrote ...
a
b
c
d
b
c
d
a
c
a
d
b
a
b
d
c
So
d
b
c
d
a
c
a
d
b
a
b
d
c
a
b
c
α
a
β
b
γ
c
Don’t remember by
Key point is that the projection of a 3D vectord onto a basis set of 3 non-coplanar vectors isUNIQUE.
a
d
c
b
Question Use the quadruple vector product to express the vector
d
1] in terms of the vectors
a
b
and
c
Answer
d
b
c
d
a
c
a
d
b
a
b
d
c
a
b
c
So, grinding away at the determinants, we find
a
b
c
18 and (
b
c
d
c
a
d
12 and (
a
b
d
So
d
a
b
c
a
b
c
Vector
p
from
c
to ANY line point
r
is
p
r
c
a
λ
ˆb
c
a
c
λ
ˆb
which has length squared
p
2
a
c
2
λ
2
λ
a
c
ˆb
λ
b
a
r
c
r−c
Easier to minimize
p
2
rather than
p
itself.
d
dλ
p
2
when
λ
a
c
ˆb
So the minimum length vector is
p
a
c
a
c
ˆb
ˆb
No surprise! It’s the component of (
a
c
perpendicular
to
ˆb
We could therefore write using the “book” formula ...
p
ˆb
a
c
ˆb
p
min
ˆb
a
c
b
a
c
ˆb
Shortest distance from point to line is along the perp line
shortest distance between two straight lines is along mutual perpendicular.
The lines are: r
a
λ
ˆb
r
c
μ
ˆd
The unit vector along the mutual perp is
ˆp
b
ˆd
b
ˆd
(Yes! Don’t forget that
ˆb
ˆd
is NOT a unit vector.)
a
c
λ
b
Q
P
μ
d
The minimum length is therefore the component of (
a
c
) in this direction
p
min
a
c
ˆb
ˆd
ˆb
ˆd
Answer Pipes A and B have axes:
r
A
λ
′
λ
r
B
μ
′
μ
(Non-unit) perpendicular to both their axes is
p
ı
ˆk
The length of the mutual perpendicular is mod
a
b
Sum of the radii of the pipes is 0
Hence the pipes do not intersect.
b
and
c
non-parallel, and
a
is a point on the plane, then
r
a
λ
b
μ
c
where
λ, μ
are scalar parameters.
a
r
b
c
NB that these are parallel to the plane, notnecessarily in the plane
O
Unit normal to the plane is
ˆn
, and a point
in the plane is
a
r
n
a
ˆn
a
^ n
r
Notice that
is the perpendicular distance to the plane from the origin.
Why not just
The plane is
r
ˆn
a
ˆn
The shortest distance
d
min
from any point to the plane is along the perpendicular.
So, the shortest distance from the
origin
to the plane is
d
min
a
ˆn
a
b
c
b
c
Now, the shortest distance from point
d
to the plane ...
d
λ
ˆn
must be a point on plane
d
λ
n
ˆn
λ
d
ˆn
d
min
λ
d
ˆn
^ n
r
d
