X Y Z Space

In this video I introduce the three axes (dimensions) in space. I introduce the concept of a point, line, plane and volume as required for Leaving Cert physi. Position Cartesian coordinates (x,y,z) are an easy and natural means of representing a position in 3D space But there are many other representations such as spherical.

In special relativity, the Minkowski spacetime is a four-dimensional manifold, created by Hermann Minkowski. It has four dimensions: three dimensions of space (x, y, z) and one dimension of time. Minkowski spacetime has a metric signature of (-+++), and describes a flat surface when no mass is present. The convention in this article is to call Minkowski spacetime simply spacetime.

However, Minkowski spacetime only applies in special relativity. General relativity used the notion of curved spacetime to describe the effects of gravity and accelerated motion.

Definition(s)[change change source]

Mathematical[change change source]

Spacetime can be thought of as a four-dimensional coordinate system in which the axes are given by

${\displaystyle (ct,x,y,z)}$

They can also be denoted by

${\displaystyle (x_1,x_2,x_3,x_4)}$

Where ${\displaystyle x_1}$ represents ${\displaystyle ct}$. The reason for measuring time in units of the speed of light times the time coordinate is so that the units for time are the same as the units for space. Spacetime has the differential for arc length given by

${\displaystyle d{s}^2=-c^{2}d{t}^2+d{x}^2+d{y}^2+d{z}^2}$

This implies that spacetime has a metric tensor given by

As before stated, spacetime is flat everywhere; to some extent, it can be thought of as a plane.

Simple[change change source]

Spacetime can be thought of as the 'arena' in which all of the events in the universe take place. All that one needs to specify a point in spacetime is a certain time and a typical spatial orientation. It is hard (virtually impossible) to visualize four dimensions, but some analogy can be made, using the method below.

Spacetime diagrams[change change source]

Hermann Minkowski introduced a certain method for graphing coordinate systems in Minkowski spacetime. As seen to the right, different coordinate systems will disagree on an object's spatial orientation and/or position in time. As you can see from the diagram, there is only one spatial axis (the x-axis) and one time axis (the ct-axis). If need be, one can introduce an extra spatial dimension, (the y-axis); unfortunately, this is the limit to the number of dimensions: graphing in four dimensions is impossible. The rule for graphing in Minkowski spacetime goes as follows:

1) The angle between the x-axis and the x'-axis is given by ${\displaystyle tan(\alpha )=\frac{v}{c}}$ where v is the velocity of the object

2) The speed of light through spacetime always makes an angle of 45 degrees with either axis.

Spacetime in general relativity[change change source]

In the general theory of relativity, Einstein used the equation

${\displaystyle R_{uv}-\frac{1}{2}{g}_{uv}R=8\pi T_{uv}}$

To allow for spacetime to actually curve; the resulting effects are those of gravity.

Related pages[change change source]

On this page...

Magnitude of a 3-D Vector
Adding 3-D Vectors
Dot Product of 3-D Vectors
Direction Cosines
Angle Between Vectors
Application

We saw earlier how to represent 2-dimensional vectors on the x-y plane.

Now we extend the idea to represent 3-dimensional vectors using the x-y-z axes. (See The 3-dimensional Co-ordinate System for background on this).

Example

The vector OP has initial point at the origin O (0, 0, 0) and terminal point at P (2, 3, 5). We can draw the vectorOP as follows:

Magnitude of a 3-Dimensional Vector

We saw earlier that the distance between 2 points in 3-dimensional space is

`'distance' AB = ` `sqrt ((x_2-x_1)^2+ (y_2-y_1)^2+ (z_2-z_1)^2)`

For the vector OP above, the magnitude of the vector is given by:

` OP = sqrt(2^2+ 3^2+ 5^2) = 6.16 'units' `

Adding 3-dimensional Vectors

Earlier we saw how to add 2-dimensional vectors. We now extend the idea for 3-dimensional vectors.

We simply add the i components together, then the j components and finally, the k components.

Example 1

Two anchors are holding a ship in place and their forces acting on the ship are represented by vectors A and B as follows:

A = 2i + 5j − 4k and B = −2i − 3j − 5k

If we were to replace the 2 anchors with 1 single anchor, what vector represents that single vector?

Answer

The problem just requires us to add the vectors to get the single resultant vector.

Dot Product of 3-dimensional Vectors

To find the dot product (or scalar product) of 3-dimensional vectors, we just extend the ideas from the dot product in 2 dimensions that we met earlier.

Example 2 - Dot Product Using Magnitude and Angle

Find the dot product of the vectors P and Q given that the angle between the two vectors is 35° and

P = 25 units and Q = 4 units

Answer

Using our formula from before for dot product:

P • Q = P Q cos θ

we have:

Example 3 - Dot Product if Vectors are Multiples of Unit Vectors

Find the dot product of the vectors A and B (these come from our anchor example above):

A = 2i + 5j − 4k and B = −2i − 3j − 5k

Answer

Direction Cosines

Suppose we have a vector OA with initial point at the origin and terminal point at A.

Suppose also that we have a unit vector in the same direction as OA. (Go here for a reminder on unit vectors).

Let our unit vector be:

u = u₁i + u₂j + u₃k

On the graph, u is the unit vector (in black) pointing in the same direction as vector OA, and i, j, and k (the unit vectors in the x-, y- and z-directions respectively) are marked in green.

We now zoom in on the vector u, and change orientation slightly, as follows:

Now, if in the diagram above,

α is the angle between u and the x-axis (in dark red),
β is the angle between u and the y-axis (in green) and
γ is the angle between u and the z-axis (in pink),

then we can use the scalar product and write:

So we can write our unit vector u as:

u = cos αi + cos βj + cos γk

These 3 cosines are called the direction cosines.

Angle Between 3-Dimensional Vectors

Earlier, we saw how to find the angle between 2-dimensional vectors. We use the same formula for 3-dimensional vectors:

`theta=arccos((P * Q)/( P Q ))`

Example 4

Find the angle between the vectors P = 4i + 0j + 7k and Q = -2i + j + 3k.

Answer

The vectors P and Q are as follows. Vector P is on the x-z plane (note that the y-value for vector P is `0`) , while Q is 'behind' the y-z plane.

Using the formula

`theta=arccos((P*Q)/( P Q ))`

we have:

And now for the denominator:

` P Q = sqrt (4^2+ (0)^2+ 7^2)` `xxsqrt((-2)^2+1^2+3^2) `

`= sqrt (65)sqrt(14)`

`=30.166 'units'`

So

θ = arccos(13 ÷ 30.166)

Therefore the angle between the vectors P and Q is

θ = 64.47°

Exercise

Find the angle between the vectors P = 3i + 4j − 7k and Q = -2i + j + 3k.

Answer

Using the formula

`theta=arccos((P*Q)/( P Q ))`

we find the dot product first:

And now for the denominator:

So

θ = arccos(−23 ÷ 32.187)

Therefore the angle between the vectors P and Q is

θ = 135.6°

Application

We have a cube ABCO PQRS which has a string along the cube's diagonal B to S and another along the other diagonal C to P

What is the angle between the 2 strings?

Answer

For convenience, we will assume that we have a unit cube (each side has length 1 unit) and we place it such that one corner of the cube is at the origin.

The unit vectors i, j, and k act in the x-, y-, and z-directions respectively. So in our diagram, since we have a unit cube,

OA = i
OC = j
OS = k

From the diagram, we see that to move from B to S, we need to go −1 unit in the x direction, −1 unit in the y-direction and up 1 unit in the z-direction. Since we have a unit cube, we can write:

BS = −i − j + k

and similarly:

CP = i − j + k

The scalar product for the vectors BS and CP is:

BS • CP = BS CP cos θ

where θ is the angle between BS and CP.

So the angle θ is given by

θ= arccos[ (BS • CP)÷ ( BS CP ) ]

Now,

BS • CP

= (−i − j + k) • (i − j + k)

= −1 + 1 + 1

= 1

and

BS CP

`= sqrt((-1)^2 + (-1)^2 + 1^2)` ` × sqrt(1^2 + (-1)^2 + 1^2)`

`= (sqrt3)(sqrt3)`

= 3

So

`θ = arccos (1/3)`

θ = 70.5°

X Y Z Access

So the angle between the strings is `70.5°`. (In this situation we assume 'angle' refers to the acute angle between the strings.)

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