__3D Vectors__

All 3D vectors can be represented by a directed line segment in 3D space R^{3}, which has a start point and an end point. This gives each vector a magnitude (the length of the line segment) and direction (from the start point to the end point). They have a rigorous definition in terms of vector spaces, but this'll do for now.

If we take the origin as out starting point we can describe any vector, **a**, by specifying its end point in cartesian coordinates (x, y, z). So any vector can be described by an ordered 3-tuple (x, y, z) where x, y and z are real numbers. In n dimensional space we could describe any vector by an ordered n-tuple (x_{1}, x_{2}, ... , x_{n}) but we'll stay in 3D.

*Definition:* The **position vector** of a point A is the vector represented by the line segment from the origin to A, and is written __a__.

__Addition of vectors__

Given the points A, B, C, let the __u__ be the vector from A to B, and __v__ be the vector from B to C. The vector __u__ + __v__ is the vector from point A to point C. On paper, if

**u **= (a, b, c) and

**v** = (x, y, z) then

__u__ + __v__ = (a + x, b + y, c + z)

__Multiplication of a vector with a scalar__

Multiplying a vector by a scalar leaves the direction of line segment unchanged, but multiplies the length by the scalar. Again, if l

is a real number and

**u **= (a, b, c) then

l **u **= (la, lb, lc)

**i**, **j**, and **k** - __The cartesian basis vectors__

__i__ is the vector from (0,0,0) to (1,0,0)

__j__ is the vector from (0,0,0) to (0,1,0)

__k__ is the vector from (0,0,0) to (0,0,1)

These vectors are the Cartesian vectors which form a basis of R^{3}. This means that any 3D vector __a__ from (0,0,0) to (x, y, z) can be written as a linear combination of **i**, **j**, and **k** like so:

__a__ = x__i__ + y__j__ + z__k__

To get used to visualizing 3D vectors try testing out this applet

__The magnitude of a vector__

Let __a__ = x__i__ + y__j__ + z__k__

then the magnitude of __a__, written a (or sometimes |__a__|), is given by:

a^{2} = x^{2} + y^{2} + z^{2}

*Proof:*

By Pythagoras' theorem

w^{2} = x^{2} + y^{2} and

a^{2} = w^{2} + z^{2}

=> a^{2} = x^{2} + y^{2} + z^{2}

__Proof of Pythagoras' theorem__

Let x be the area of the larger square, then

x = (a + b)^{2} but also

x = c^{2} + 4 (1/2 ab )

=> (a + b)^{2} = c^{2} + 2ab

=> a^{2} + b^{2} + 2ab = c^{2} + 2ab

=> a^{2} + b^{2} = c^{2}

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