Three-Dimensional Coordinate Systems

Three-Dimensional Coordinates and Vectors

Basic definition

Two-dimensional Cartesian coordinates as we've discussed so far can be easily extended to three-dimensions by adding one more value: $z$ . If the standard $(x,y)$ coordinate axes are drawn on a sheet of paper, the $z$-axis would extend upwards off of the paper.

Similar to the two coordinate axes in two-dimensional coordinates, there are three coordinate planes in space. These are the $xy$-plane, the $yz$-plane, and the $xz$-plane. Each plane is the "sheet of paper" that contains both axes the name mentions. For instance, the $yz$-plane contains both the $y$ and $z$ axes and is perpendicular to the $x$-axis.

Therefore, vectors can be extended to three dimensions by simply adding the $z$ value.

$\mathbf {u} ={\begin{pmatrix}x\\y\\z\end{pmatrix}}$

To facilitate standard form notation, we add another standard unit vector:

$\mathbf {k} ={\begin{pmatrix}0\\0\\1\end{pmatrix}}$

Again, both forms (component and standard) are equivalent.

${\begin{pmatrix}1\\2\\3\end{pmatrix}}=1\mathbf {i} +2\mathbf {j} +3\mathbf {k}$

Magnitude: Magnitude in three dimensions is the same as in two dimensions, with the addition of a $z$ term in the radicand.

$|\mathbf {u} |={\sqrt {u_{x}^{2}+u_{y}^{2}+u_{z}^{2}}}$

Three dimensions

The polar coordinate system is extended into three dimensions with two different coordinate systems, the cylindrical and spherical coordinate systems, both of which include two-dimensional or planar polar coordinates as a subset. In essence, the cylindrical coordinate system extends polar coordinates by adding an additional distance coordinate, while the spherical system instead adds an additional angular coordinate.

Cylindrical coordinates

a point plotted with cylindrical coordinates

The cylindrical coordinate system is a coordinate system that essentially extends the two-dimensional polar coordinate system by adding a third coordinate measuring the height of a point above the plane, similar to the way in which the Cartesian coordinate system is extended into three dimensions. The third coordinate is usually denoted $h$ , making the three cylindrical coordinates $(r,\theta ,h)$ .

The three cylindrical coordinates can be converted to Cartesian coordinates by

${\begin{aligned}x&=r\cos(\theta )\\y&=r\sin(\theta )\\z&=h\end{aligned}}$

Spherical coordinates

A point plotted using spherical coordinates

Polar coordinates can also be extended into three dimensions using the coordinates $(\rho ,\phi ,\theta )$ , where $\rho$ is the distance from the origin, $\phi$ is the angle from the $z$-axis (called the colatitude or zenith and measured from 0 to 180°) and $\theta$ is the angle from the $x$-axis (as in the polar coordinates). This coordinate system, called the spherical coordinate system, is similar to the latitude and longitude system used for Earth, with the origin in the centre of Earth, the latitude $\delta$ being the complement of $\phi$ , determined by $\delta =90^{\circ }-\phi$ , and the longitude $l$ being measured by $l=\theta -180^{\circ }$ .

The three spherical coordinates are converted to Cartesian coordinates by

${\begin{aligned}x&=\rho \sin(\phi )\cos(\theta )\\y&=\rho \sin(\phi )\sin(\theta )\\z&=\rho \cos(\phi )\end{aligned}}$

$\rho ={\sqrt {x^{2}+y^{2}+z^{2}}}$

$\theta =\arctan \left({\frac {y}{x}}\right)$

$\phi =\arccos \left({\frac {z}{\rho }}\right)$

Cross Product

The cross product of two vectors is a determinant:

$\mathbf {u} \times \mathbf {v} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\u_{x}&u_{y}&u_{z}\\v_{x}&v_{y}&v_{z}\end{vmatrix}}$

and is also a pseudovector.

The cross product of two vectors is orthogonal to both vectors. The magnitude of the cross product is the product of the magnitude of the vectors and the sin of the angle between them.

$|\mathbf {u} \times \mathbf {v} |=|\mathbf {u} ||\mathbf {v} |\sin(\theta )$

This magnitude is the area of the parallelogram defined by the two vectors.

The cross product is linear and anticommutative. For any numbers $a,b$ ,

$\mathbf {u} \times \left(a\mathbf {v} +b\mathbf {w} \right)=a\mathbf {u} \times \mathbf {v} +b\mathbf {u} \times \mathbf {w} \qquad \mathbf {u} \times \mathbf {v} =-\mathbf {v} \times \mathbf {u}$

If both vectors point in the same direction, their cross product is 0.

Deriving the Cross Product

A depiction of the cross product of vectors $\mathbf {u}$ and $\mathbf {v}$.

Start with the following definition for the cross product: $\mathbf {u} \times \mathbf {v} =(|\mathbf {u} ||\mathbf {v} |\sin \theta ){\hat {\mathbf {n} }}$. Vector ${\hat {\mathbf {n} }}$ is a unit length vector that is perpendicular to both $\mathbf {u}$ and $\mathbf {v}$ and oriented according to the "right hand rule". Angle $\theta$ is the counterclockwise angle from $\mathbf {u}$ to $\mathbf {v}$ within the plane that is orthogonal to ${\hat {\mathbf {n} }}$.

The formula $\mathbf {u} \times \mathbf {v} =(u_{y}v_{z}-u_{z}v_{y})\mathbf {i} +(u_{z}v_{x}-u_{x}v_{z})\mathbf {j} +(u_{x}v_{y}-u_{y}v_{x})\mathbf {k}$ can be derived by exploiting the bilinearity of the cross product. To establish the cross product as a bilinear operator, the following must be established:

1. Holding $\mathbf {u}$ constant, the cross product $\mathbf {u} \times \mathbf {v}$ must be linear with respect to $\mathbf {v}$.
2. Holding $\mathbf {v}$ constant, the cross product $\mathbf {u} \times \mathbf {v}$ must be linear with respect to $\mathbf {u}$.

From the definition $\mathbf {u} \times \mathbf {v} =(|\mathbf {u} ||\mathbf {v} |\sin \theta ){\hat {\mathbf {n} }}$, the anticommutative property of the cross product can be inferred: $\mathbf {u} \times \mathbf {v} =-\mathbf {v} \times \mathbf {u}$ (vector ${\hat {\mathbf {n} }}$ reverses direction when $\mathbf {u}$ and $\mathbf {v}$ are swapped). This means that linearity with respect to $\mathbf {v}$ implies linearity with respect to $\mathbf {u}$. It is hence only necessary to establish that the cross product is linear with respect to $\mathbf {v}$ to establish bilinearity.

The perpendicular component of $\mathbf {v}$ relative to the line $L(\mathbf {u} )$.

$\mathbf {u} \times \mathbf {v} =(|\mathbf {u} ||\mathbf {v} |\sin \theta ){\hat {\mathbf {n} }}=|\mathbf {u} |{\textbf {rotate}}({\textbf {perp}}(\mathbf {v} |\mathbf {u} )|\mathbf {u} )$ where:

• ${\textbf {perp}}(\mathbf {v} |\mathbf {u} )$ is the component of $\mathbf {v}$ that is perpendicular to a line $L(\mathbf {u} )$ whose direction is the direction of $\mathbf {u}$. ${\textbf {perp}}(\mathbf {v} |\mathbf {u} )=(|\mathbf {v} |\sin \theta ){\hat {\mathbf {m} }}$ where vector ${\hat {\mathbf {m} }}$ is a unit length vector that is parallel to ${\textbf {perp}}(\mathbf {v} |\mathbf {u} )$.
• ${\textbf {rotate}}(\mathbf {w} |\mathbf {u} )$ is a 90 degree counterclockwise rotation of $\mathbf {w}$ around $L(\mathbf {u} )$. Note that ${\textbf {rotate}}({\hat {\mathbf {m} }}|\mathbf {u} )={\hat {\mathbf {n} }}$.

It can easily be observed that ${\textbf {perp}}(\mathbf {v} |\mathbf {u} )$ is linear with respect to $\mathbf {v}$ with $\mathbf {u}$ held constant, and that ${\textbf {rotate}}(\mathbf {w} |\mathbf {u} )$ is linear with respect to $\mathbf {w}$ with $\mathbf {u}$ held constant. The cross product $\mathbf {u} \times \mathbf {v} =(|\mathbf {u} ||\mathbf {v} |\sin \theta ){\hat {\mathbf {n} }}=|\mathbf {u} |{\textbf {rotate}}({\textbf {perp}}(\mathbf {v} |\mathbf {u} )|\mathbf {u} )$ is linear with respect to $\mathbf {v}$, and therefore the cross product is a bilinear operator.

The bilinearity of the cross product now enables the derivation:

$\mathbf {u} \times \mathbf {v} =(u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )\times (v_{x}\mathbf {i} +v_{y}\mathbf {j} +v_{z}\mathbf {k} )$ $=v_{x}((u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )\times \mathbf {i} )+v_{y}((u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )\times \mathbf {j} )+v_{z}((u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )\times \mathbf {k} )$ $=u_{x}v_{x}(\mathbf {i} \times \mathbf {i} )+u_{y}v_{x}(\mathbf {j} \times \mathbf {i} )+u_{z}v_{x}(\mathbf {k} \times \mathbf {i} )+u_{x}v_{y}(\mathbf {i} \times \mathbf {j} )+u_{y}v_{y}(\mathbf {j} \times \mathbf {j} )+u_{z}v_{y}(\mathbf {k} \times \mathbf {j} )+u_{x}v_{z}(\mathbf {i} \times \mathbf {k} )+u_{y}v_{z}(\mathbf {j} \times \mathbf {k} )+u_{z}v_{z}(\mathbf {k} \times \mathbf {k} )$ $=u_{x}v_{y}\mathbf {0} +u_{y}v_{x}(-\mathbf {k} )+u_{z}v_{x}\mathbf {j} +u_{x}v_{y}\mathbf {k} +u_{y}v_{y}\mathbf {0} +u_{z}v_{y}(-\mathbf {i} )+u_{x}v_{z}(-\mathbf {j} )+u_{y}v_{z}\mathbf {i} +u_{z}v_{z}\mathbf {0}$ $=(u_{y}v_{z}-u_{z}v_{y})\mathbf {i} +(u_{z}v_{x}-u_{x}v_{z})\mathbf {j} +(u_{x}v_{y}-u_{y}v_{x})\mathbf {k}$

Therefore $\mathbf {u} \times \mathbf {v} =(u_{y}v_{z}-u_{z}v_{y})\mathbf {i} +(u_{z}v_{x}-u_{x}v_{z})\mathbf {j} +(u_{x}v_{y}-u_{y}v_{x})\mathbf {k}$.

Triple Products

If we have three vectors we can combine them in two ways, a triple scalar product,

$\mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )$

and a triple vector product

$\mathbf {u} \times (\mathbf {v} \times \mathbf {w} )$

The triple scalar product is a determinant

$\mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )={\begin{vmatrix}u_{x}&u_{y}&u_{z}\\v_{x}&v_{y}&v_{z}\\w_{x}&w_{y}&w_{z}\end{vmatrix}}$

If the three vectors are listed clockwise, looking from the origin, the sign of this product is positive. If they are listed anticlockwise the sign is negative.

The order of the cross and dot products doesn't matter.

$\mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )=(\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w}$

Either way, the absolute value of this product is the volume of the parallelepiped defined by the three vectors $\mathbf {u} ,\mathbf {v} ,\mathbf {w}$

The triple vector product can be simplified

$\mathbf {u} \times (\mathbf {v} \times \mathbf {w} )=(\mathbf {u} \cdot \mathbf {w} )\mathbf {v} -(\mathbf {u} \cdot \mathbf {v} )\mathbf {w}$

This form is easier to do calculations with.

The triple vector product is not associative.

$\mathbf {u} \times (\mathbf {v} \times \mathbf {w} )\neq (\mathbf {u} \times \mathbf {v} )\times \mathbf {w}$

There are special cases where the two sides are equal, but in general the brackets matter. They must not be omitted.

Three-Dimensional Lines and Planes

We will use $\mathbf {r}$ to denote the position of a point.

The multiples of a vector, $\mathbf {a}$ all lie on a line through the origin. Adding a constant vector $\mathbf {b}$ will shift the line, but leave it straight, so the equation of a line is,

$\mathbf {r} =s\mathbf {a} +\mathbf {b}$

This is a parametric equation. The position is specified in terms of the parameter $s$ .

Any linear combination of two vectors, $\mathbf {a} ,\mathbf {b}$ lies on a single plane through the origin, provided the two vectors are not colinear. We can shift this plane by a constant vector again and write

$\mathbf {r} =s\mathbf {a} +t\mathbf {b} +\mathbf {c}$

If we choose $\mathbf {a} ,\mathbf {b}$ to be orthonormal vectors in the plane (i.e. unit vectors at right angles) then $s,t$ are Cartesian coordinates for points in the plane.

These parametric equations can be extended to higher dimensions.

Instead of giving parametric equations for the line and plane, we could use constraints. E.g., for any point in the $xy$ plane $z=0$

For a plane through the origin, the single vector normal to the plane, $\mathbf {n}$ , is at right angle with every vector in the plane, by definition, so

$\mathbf {r} \cdot \mathbf {n} =0$

is a plane through the origin, normal to $\mathbf {n}$ .

For planes not through the origin we get

$(\mathbf {r} -\mathbf {a} )\cdot \mathbf {n} =0\qquad \mathbf {r} \cdot \mathbf {n} =\mathbf {a} \cdot \mathbf {n}$

A line lies on the intersection of two planes, so it must obey the constraint for both planes, i.e.

$\mathbf {r} \cdot \mathbf {n} =a\qquad \mathbf {r} \cdot \mathbf {m} =b$

These constraint equations con also be extended to higher dimensions.

Resources

• Plotting Points In a Three Dimensional Coordinate System Video by The Organic Chemistry Tutor 2017

Licensing

Content obtained and/or adapted from:

• Vectors, Wikibooks: Calculus under a CC BY-SA license