Vectors

A vector $\vec{v} \in \mathbb{R}^n$ is an $n$-tuple of real numbers. For example, consider a vector that has three components:

$$ \vec{v} = (v_1,v_2,v_3) \ \in \ (\mathbb{R},\mathbb{R},\mathbb{R}) \equiv \mathbb{R}^3. $$

To specify the vector $\vec{v}$, we specify the values for its three components $v_1$, $v_2$, and $v_3$.

A matrix $A \in \mathbb{R}^{m\times n}$ is a rectangular array of real numbers with $m$ rows and $n$ columns. A vector is a special type of matrix; we can think of a vector $\vec{v}\in \mathbb{R}^n$ either as a row vector ($1\times n$ matrix) or a column vector ($n \times 1$ matrix). Because of this equivalence between vectors and matrices, there is no need for a special vector object in SymPy, and Matrix objects are used for vectors as well.

This is how we define vectors and compute their properties:

u = Matrix([[4,5,6]])  # a row vector = 1x3 matrix
v = Matrix([[7],
            [8],       # a col vector = 3x1 matrix 
            [9]])

v.T                    # use the transpose operation to convert a col vec to a row vec

$$\left[\begin{matrix}7 & 8 & 9\end{matrix}\right]$$

u[0]                   # 0-based indexing for entries

$$4$$

u.norm()               # length of u

$$\sqrt{77}$$

uhat = u/u.norm()      # unit-length vec in same dir as u
uhat

$$\left[\begin{matrix}\frac{4 \sqrt{77}}{77} & \frac{5 \sqrt{77}}{77} & \frac{6 \sqrt{77}}{77}\end{matrix}\right]$$

uhat.norm()

$$1$$

Dot product

The dot product of the 3-vectors $\vec{u}$ and $\vec{v}$ can be defined two ways:

$$ \vec{u}\cdot\vec{v} \equiv \underbrace{u_xv_x+u_yv_y+u_zv_z}_{\textrm{algebraic def.}} \equiv \underbrace{\|\vec{u}\|\|\vec{v}\|\cos(\varphi)}_{\textrm{geometric def.}} \quad \in \mathbb{R}, $$

where $\varphi$ is the angle between the vectors $\vec{u}$ and $\vec{v}$. In SymPy,

u = Matrix([ 4,5,6])
v = Matrix([-1,1,2])
u.dot(v)

$$13$$

We can combine the algebraic and geometric formulas for the dot product to obtain the cosine of the angle between the vectors

$$ \cos(\varphi) = \frac{ \vec{u}\cdot\vec{v} }{ \|\vec{u}\|\|\vec{v}\| } = \frac{ u_xv_x+u_yv_y+u_zv_z }{ \|\vec{u}\|\|\vec{v}\| }, $$

and use the acos function to find the angle measure:

acos(u.dot(v)/(u.norm()*v.norm())).evalf()  # in radians = 52.76 degrees

$$0.921263115666387$$

Just by looking at the coordinates of the vectors $\vec{u}$ and $\vec{v}$, it's difficult to determine their relative direction. Thanks to the dot product, however, we know the angle between the vectors is $52.76^\circ$, which means they kind of point in the same direction. Vectors that are at an angle $\varphi=90^\circ$ are called orthogonal, meaning at right angles with each other. The dot product of vectors for which $\varphi > 90^\circ$ is negative because they point mostly in opposite directions.

The notion of the “angle between vectors” applies more generally to vectors with any number of dimensions. The dot product for $n$-dimensional vectors is $\vec{u}\cdot\vec{v}=\sum_{i=1}^n u_iv_i$. This means we can talk about “the angle between” 1000-dimensional vectors. That's pretty crazy if you think about it—there is no way we could possibly “visualize” 1000-dimensional vectors, yet given two such vectors we can tell if they point mostly in the same direction, in perpendicular directions, or mostly in opposite directions.

The dot product is a commutative operation $\vec{u}\cdot\vec{v} = \vec{v}\cdot\vec{u}$:

u.dot(v) == v.dot(u)

True

Projections

Dot products are used for computing projections. Assume you're given two vectors $\vec{u}$ and $\vec{n}$ and you want to find the component of $\vec{u}$ that points in the $\vec{n}$ direction. The following formula based on the dot product will give you the answer:

$$ \Pi_{\vec{n}}( \vec{u} ) \equiv \frac{ \vec{u} \cdot \vec{n} }{ \| \vec{n} \|^2 } \vec{n}. $$

This is how to implement this formula in SymPy:

u = Matrix([4,5,6])
n = Matrix([1,1,1])
(u.dot(n) / n.norm()**2)*n  # projection of v in the n dir

$$\left[\begin{matrix}5\\5\\5\end{matrix}\right]$$

In the case where the direction vector $\hat{n}$ is of unit length $\|\hat{n}\| = 1$, the projection formula simplifies to $\Pi_{\hat{n}}( \vec{u} ) \equiv (\vec{u}\cdot\hat{n})\hat{n}$.

Consider now the plane $P$ defined by $(1,1,1)\cdot[(x,y,z)-(0,0,0)]=0$. A plane is a two dimensional subspace of $\mathbb{R}^3$. We can decompose any vector $\vec{u} \in \mathbb{R}^3$ into two parts $\vec{u}=\vec{v} + \vec{w}$ such that $\vec{v}$ lies inside the plane and $\vec{w}$ is perpendicular to the plane (parallel to $\vec{n}=(1,1,1)$).

To obtain the perpendicular-to-$P$ component of $\vec{u}$, compute the projection of $\vec{u}$ in the direction $\vec{n}$:

w = (u.dot(n) / n.norm()**2)*n
w

$$\left[\begin{matrix}5\\5\\5\end{matrix}\right]$$

To obtain the in-the-plane-$P$ component of $\vec{u}$, start with $\vec{u}$ and subtract the perpendicular-to-$P$ part:

v = u - (u.dot(n)/n.norm()**2)*n  # same as u - w
v

$$\left[\begin{matrix}-1\\0\\1\end{matrix}\right]$$

You should check on your own that $\vec{v}+\vec{w}=\vec{u}$ as claimed.

Cross product

The cross product, denoted $\times$, takes two vectors as inputs and produces a vector as output. The cross products of individual basis elements are defined as follows:

$$ \hat{\imath}\times\hat{\jmath} =\hat{k}, \qquad \hat{\jmath}\times\hat{k} =\hat{\imath}, \qquad \hat{k}\times \hat{\imath}= \hat{\jmath}. $$

Here is how to compute the cross product of two vectors in SymPy:

u = Matrix([ 4,5,6])
v = Matrix([-1,1,2])
u.cross(v)

$$\left[\begin{matrix}4\\-14\\9\end{matrix}\right]$$

The vector $\vec{u}\times \vec{v}$ is orthogonal to both $\vec{u}$ and $\vec{v}$. The norm of the cross product $\|\vec{u}\times \vec{v}\|$ is proportional to the lengths of the vectors and the sine of the angle between them:

(u.cross(v).norm()/(u.norm()*v.norm())).n()

$$0.796366206088088$$

The name “cross product” is well-suited for this operation since it is calculated by “cross-multiplying” the coefficients of the vectors:

$$ \vec{u}\times\vec{v}= \left( u_yv_z-u_zv_y, \ u_zv_x-u_xv_z, \ u_xv_y-u_yv_x \right). $$

By defining individual symbols for the entries of two vectors, we can make SymPy show us the cross-product formula:

u1,u2,u3 = symbols('u1:4')
v1,v2,v3 = symbols('v1:4')
Matrix([u1,u2,u3]).cross(Matrix([v1,v2,v3]))

$$\left[\begin{matrix}u_{2} v_{3} - u_{3} v_{2}\\- u_{1} v_{3} + u_{3} v_{1}\\u_{1} v_{2} - u_{2} v_{1}\end{matrix}\right]$$

The dot product is anti-commutative $\vec{u}\times\vec{v} = -\vec{v}\times\vec{u}$:

u.cross(v)

$$\left[\begin{matrix}4\\-14\\9\end{matrix}\right]$$

v.cross(u)

$$\left[\begin{matrix}-4\\14\\-9\end{matrix}\right]$$

The product of two numbers and the dot product of two vectors are commutative operations. The cross product, however, is not commutative: $\vec{u}\times\vec{v} \neq \vec{v}\times\vec{u}$.