Part 2: Vectors in 3 dimensions

Three dimensional space, which is denoted by R³, is often modeled as a horizontal xy-plane with a vertical z-axis intersecting the plane at the origin. 

We assume that positive z-coordinates are above the xy-plane.  This means that the coordinate system is right-handed because if the fingers of the right hand are wrapped about the z-axis in the counter-clockwise direction, then the thumb of the right hand points in the positive z-direction.

The point P₁(x₁,y₁,z₁) in R³ is the point that has a vertical displacement of z₁ above the point (x₁,y₁,0) in the xy-plane.  Often we write a point (x₁,y₁,0) as (x₁,y₁) .

Given points P₁( x₁,y₁,z₁) and P₂(x₂,y₂,z₂) in R³, we define the vector between them to be

v = P1P2  = áx2-x1,y2-y1,z2-z1 ñ

(2)

Notice that in 3 dimensions a vector is of the form v = áa,b,c ñ where a is a run, b is a shift left or right, and c is a rise. Thus, vectors generalize to 3 dimensions whereas slopes do not. 

To aid in visualization, this text will include many figures which are interactive.  These figures are prepared using the packages Javaview and LiveGraphics3D.  For example, if you "click and drag" on the figure in example 4, it will rotate. Right-clicking produces more options for interactivity.

EXAMPLE 4    Find the vector with initial point P₁(4,1,2) and terminal point P₂(1,6,5) .

Maple/Javaview Figure

Solution: To do so, we use (2) to obtain

v = á 1-4, 6-1, 5-2 ñ = á-3, 5, 3 ñ

Given vectors u = á u₁, u₂, u₃ ñ and v = á v₁, v₂, v₃ ñ, we define

u + v = á u1 + v1, u2 + v2, u3 + v3 ñ

and for a scalar k we define ku = á k u₁, k u₂, k u₃ ñ.  Two dimensional vectors v = á v₁, v₂ ñ are actually 3-dimensional vectors of the form v = áv₁, v₂, 0ñ, so that the same arithmetic applies.

The basic vectors are defined  

i = á1,0,0ñ,  j = á0,1,0ñ,  and  k = á0,0,1ñ,

We say that the basic vectors i, j, and k form a basis for R³ in that any vector á a, b, c ñ can be written as

áa, b, c ñ = ai + bj + ck

Conversely, we say that ai + bj + ck is a linear combination of i, j, and k.

EXAMPLE 5    Convert the vector  v = 2i + 4j - 3k to component form, and then sketch the vector.  

Solution: The definitions of the basis vectors lead to

2i + 4j - 3k  =  2á1,0,0ñ + 4á0,1,0ñ -  3á0,0,1ñ  =  á2,0,0ñ + á0,4,0ñ + á0,0,-3ñ  = á 2, 4, -3 ñ

Thus, v = á 2, 4, -3 ñ, which is shown with initial point at the origin in the figure below.

Maple Graphics Export

Note:  The use of i, j, and k  is very common, but other notations are also used.  For example, in some settings the basic vectors may be denoted

ex = á 1,0,0 ñ ,    ey = á 0,1,0 ñ ,    and    ez = á 0,0,1 ñ

or capital letters I, J, and K may be used in place of lower case.