1. Vectors in space - 3 dimensions
Vectors can exist in 3 dimensions (space) with axes (x,y,z). z
(4,0,0)
(0,5,0) x
(0,0,2) y

2. Distances between 2 points in space
Find the direct distance between O and C. C
(4,3,5) B
(4,3,0) O
(0,0,0) A
(4,0,0)
Verify this with the example above:
In general the distance between two points:
O(0,0,0)
C(4,3,5)

3. Distances between 2 points in space
2. A plane starts its decent to an airport from its coordinate at (1200,350,150). The runway is considered the origin. Calculate the direct distance for its descent and the angle at which the plane makes with the runway on touchdown.
1. Find the direct distance between the points A(1,-3,4) and B(3,4,9). Leave your answer as a surd.
2. P has a position vector .
Q has position vector .
Plane
150
a) Find the vector PQ.
350
Runway
1200
b) Find the direct distance between P and Q.
1250
c) M exists at the mid-point of PQ. Find the coordinates of M.
by using Pythagoras
by using Pythagoras again
by using SOHCAHTOA

4. Vector notation
Vectors do not have to be written in column form. They are often written using i,j, and k, such that:
Write each of the following column vectors in the form
xi+yj+zk. 1. 2. 3.
2i+3j+5k
can be written as 2i+4j+7k.
4i-j-3k
9i-2j

5. 3d vector operations
All the same rules that 2d vectors have apply to 3d vectors, including unit vectors, adding, subtracting, multiplying by scalars, dot product, magnitude and parallel lines.
Try each of the following based on the matrices below.
3. length of a
4. unit vector b
5. the values of y and z so that vectors b and d are parallel.
6. dot product of b and c.
a=9i-2j b=3i+5j-2k
c=-i+7j+2k d=-9i+yj+zk
1. 3a+2c
2. 5b-c
y=-15, z=6
25i+8j+4k 28
16i+18j-12k

6. Collinear points
Points are known to be collinear if they exist on the same line. C B A
Check that kAB=BC.
Show that the points A(3,1,-2), B(5,-2,2) and C(11,-11,14) are collinear and find the ratio of AB:BC.
Yes, therefore collinear.
k=3
If the points are collinear then kAB=BC.
3AB:BC