# Vector Arithmetic

Pre-requisites: Vectors
One of the differences between vectors and points is that while it doesn't
really make sense to add or subtract points, it does make sense to add
and subtract vectors. Here's how:

To add two vectors
,
we consider
as the displacement vector
for some P and then
as
the displacement vector
for some Q. The answer,
is
then the displacement vector
.

This definition of
vector arithmetic has the singular advantage that it agrees with nature:
it is a physical law that if a particle is being acted on by two
forces
,
the result is the same as if the particle were only being
acted on by the one force
.
The following picture proves that vector addition is commutative;
i.e., that
:

To define how to subtract two vectors, it suffices to define how to
negate a single vector, as
.
This is easy enough:
is simply the vector with the same magnitude as
, but with the opposite direction. To put it another way,
if
for some point P, then
.

Finally, we can define how to multiply a vector by a scalar real number:
for some non-negative real number
*c* is the vector with the same direction as
,
but with c times
's magnitude.
This definition of scalar multiplication has all of the properties you
would expect it to have:
and more generally
.

Fortunately, all of these operations are easy to calculate in Cartesian
coordinates. The trick is to realize that
(*v*_{1}, *v*_{2}, *v*_{3}) and
are the same vector. Why? Because they have the
same projections onto the coordinate axes. Once we have this knowledge,
we can see that
, or
that *coordinate-wise, vector addition is the same as tuple addition*.

It should then come as no surprise that multiplying a vector by a scalar
is the same as multiplying a tuple by a scalar:
. To prove this, let
for some point *P* and let
for some point *Q*. By hypothesis, *O*, *P*, and *Q*
all lie on the same line. If we project *P* and *Q* onto the
XY plane and call the results *P'* and *Q* respectively, then
we get two similar triangles:
. We are given that
, so by the Similar Triangles Theorem
.
But
, so the Z component of
*c* ( *v*_{1}, *v*_{2}, *v*_{3} )
must be *c* *v*_{3}. The same reasoning holds for
the X and Y components, so the result follows.

### Exercises:

- Prove that vector addition is associative; i.e, that
.

- Prove that
.

- Prove that
.

- Prove that
.

- Prove that
.

- Show that
.

- In the following window, draw in
- the sum of the red and blue vector
- the red vector minus the blue vector
- three times the blue vector

- Prove that
.

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thomasc@athena.mit.edu
Last modified 1 July 1997