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DIFFERENTIAL GEOMETRY COURSE

1.* VECTORS*

*Definition 1:* Consider points *X*, *Y*, ... in space. Let *XY* be the directed line segment from *X* to *Y*.
(We call *X* the tail and *Y* the head).
So this line segment has a length and a direction (except if *X*=*Y*).

We call two such directed line segments equivalent if they have the same length and the same direction, or if they have both length 0.
We call each equivalence class a vector (and the class of the line segments with length 0 is called the zero vector, *0*).

Now let *O* be a fixed point in space (, called origin). We associate each point *X* with a unique vector *x* with representative *OX*. We call *OX*
the main representative of *x*, and *x* the place vector of *X*.

The length of a vector *x*, which we denote by ||*x*||, is the length of any of its representatives.

*Definition 2:* If *a* is a vector and λ a real number, then λ*a* is the vector with ||λ*a*|| = |λ| ||*a*||,
having the same direction as *a* if λ > 0, and the opposite direction if λ < 0.

We add vectors using the head-on-tail method (rule of parallelograms)

*Note 3:*

(vector representation of a straight line)

The vectors *s*+λ*v* (λ ∈ ℜ) are place vectors of points on a straight line.
(We call *s* supporting vector and *v* direction vector of this line.)

Whenever *A* and *B* are arbitrary points, the place vectors of the points on the line through *A* and *B* all have the form *a*+λ(*b*-*a*).

Check the following: *XY*+*YZ*=*XZ*, so *AB*=*OB*-*OA*=*b*-*a*.

(coordinates in a plane)

Suppose that *O*, *A* and *B* don't lie on a (straight) line. Let *X* be an arbitrary point in the plane α door
*O*, *A* and *B*. Then we can uniquely write *X* in the form λ*a*+μ*b* for certain real numbers λ
and μ; we call λ and μ coordinates of *x* with respect to base *a*, *b*.

(vector representation of a plane)

In general, the vectors *s*+λ*v*+μ*w* are place vectors of points in a plane.

The place vectors of the points in the plane through three non-collinear points *A*, *B* and *C* all have the form
*a*+λ(*b*-*a*)+μ(*c*-*a*).

(coordinates in space)

Suppose that *O*, *A*, *B* don't lie *C* in one plane. Let *X* be an arbitrary point in space.
Then we can uniquely write *x* in the form λ*a*+μ*b*+ν*c* for certain real numbers λ, μ and ν (coordinates of
*x* with respect to base *a*, *b*, *c*).

If *a*, *b* and *c* have length 1 and each pair out of them is perpendicular,
we call the base orthonormal.

(natural coordinates)

We start with a fixed orthonormal base __e___{1}, __e___{2}, __e___{3}, where __e___{3} is chosen with
__e___{1} and __e___{2} according to the corkscrew rule.
We denote *x*=x_{1}__e___{1}+x_{2}__e___{2}+x_{3}__e___{3}
by *x*=(x_{1},x_{2},x_{3}).

Let *x*=(x_{1},x_{2},x_{3}) and *y*=(y_{1},y_{2},y_{3}). Then, for each real number λ, λ*x*=
(λx_{1},λx_{2},λx_{3}) and *x*+*y*=(x_{1}+y_{1},x_{2}+y_{2},x_{3}+y_{3}).

According to Pythagoras, the distance between *x* and *y* is equal to
||*x*-*y*||=√((x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+(x_{3}-y_{3})^{2}).

*Definition 4:* The inner product (scalar product) of *x* and *y* is the number
*x*.*y* := x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}.

The outer product (vector product) of *x* and *y* is the vector
*x*⊗*y* := (x_{2}y_{3}-x_{3}y_{2},x_{3}y_{1}-x_{1}y_{3},x_{1}y_{2}-x_{2}y_{1}).

*Problem 5:*Prove by using the cosine rule (from triangular geometry) that *x*.*y*=||*x*||.||*y*||.cos(φ), where φ is the angle between
*x* and *y*.

Using the inner product, prove that *x*⊗*y* is perpendicular to *x* and *y*.

*Note 6:* (equations of a plane)

Let *n* be an arbitrary vector, not equal to *0*. The points *X* whose place vectors satisfy *x*.*n*=0 are lying in the plane through *0*
perpendicular to *n*. Eeach plane with equation *x*.*n*=c (with c∈ℜ) is parallel to this plane.

*Problem 7:* Prove the identities det(*a*,*b*,*c*)=*a*⊗*b*.*c* and
*a*⊗(*b*⊗*c*)=(*a*.*c*)*b*-(*a*.*b*)*c*.

Thereafter, prove that det(*x*-*a*,*b*-*a*,*c*-*a*)=0 and
(*x*-*a*).((*b*-*a*)⊗(*c*-*a*))=0 are equations of the plane through *A*, *B* and *C*.

*Problem 8:* Suppose __b___{1}, __b___{2}, __b___{3} is an orthonormal base, and
*x*=λ_{1}__b___{1}+λ_{2}__b___{2}+λ_{3}__b___{3}. Prove that
*x*.__b___{i}=λ_{i}.

*Problem 9:* Let *x*(t)=(x_{1}(t),x_{2}(t),x_{3}(t)), where x_{1},x_{2} and x_{3} are differentiable functions of a
real parameter t that runs through an interval I. Analogously, let *y*(t)=(y_{1}(t),y_{2}(t),y_{3}(t)).

Denote x_{i}'(t) := (d/dt) x_{i}(t) and *x'*(t)=(x_{1}'(t),x_{2}'(t),x_{3}'(t)), etc.

Prove:

a) (d/dt) *x*(t).*y*(t) = *x*'(t).*y*(t)+*x*(t).*y*'(t).

b) If ||*x*(t)||=1 for t∈I, then *x*(t) is perpendicular to *x*'(t) for all t∈I.

c) (d/dt) *x*(t)⊗*y*(t) = *x*'(t)⊗*y*(t)+*x*(t)⊗*y*'(t).

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