Tutorial 3

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Vectors: position vectors, magnitude, scalar multiplication
• Unit vectors and direction cosines
• Vector, parametric and Cartesian equations of lines
• Dot product and cross product
• Gradient, divergence and curl

Introduction

A vector is just a Matrix with one column. Matrix([2, 4]) is the position vector $\langle 2, 4\rangle$ and Matrix([1, 3, -7]) is $\underset{\sim}{i}+3\underset{\sim}{j}-7\underset{\sim}{k}$.

import sympy as sy
from sympy.abc import x, y, z, t
from sympy.solvers import solve
from sympy import sqrt, tan, sin, cos, sec, pi, ln, exp
from sympy import atan, acos, asin, diff, Matrix, simplify, nsimplify
sy.init_printing(use_latex=True)

# --- handy helpers used in Q7, Q8 and Q9 ---
def gradient(f, vars=(x, y, z)):
    """Gradient of a scalar field, returned as a column vector."""
    return Matrix([diff(f, v) for v in vars])

def divergence(F, vars=(x, y, z)):
    """Divergence of a vector field given as a list/Matrix of components."""
    return sum(diff(F[i], vars[i]) for i in range(len(vars)))

def curl(F):
    """Curl of a 3D vector field given as a list/Matrix of components."""
    return Matrix([diff(F[2], y) - diff(F[1], z),
                   diff(F[0], z) - diff(F[2], x),
                   diff(F[1], x) - diff(F[0], y)])

Question 1

Sketch $P(3,-1,5)$ and $Q(2,1,-1)$ and find the distance between them.

P = Matrix([3, -1, 5])
Q = Matrix([2, 1, -1])
PQ = Q - P
display(PQ, sqrt(PQ.dot(PQ)))      # <-1,2,-6>, sqrt(41)

The distance is $\sqrt{1^{2}+(-2)^{2}+(-6)^{2}}=\sqrt{41}$.

Question 2

For $\underset{\sim}{a}=\langle 2,4\rangle$, compute $3\underset{\sim}{a}$, $\frac{1}{2}\underset{\sim}{a}$ and $-2\underset{\sim}{a}$.

a = Matrix([2, 4])
display(3*a)          # <6, 12>
display(a/2)          # <1, 2>
display(-2*a)         # <-4, -8>

Scalar multiplication never changes the direction of a vector, only its length — except for a negative scalar, which reverses the direction. A scalar larger than $1$ stretches the vector, a scalar between $0$ and $1$ shrinks it, and $-2$ doubles the length while flipping it through $180^\circ$.

Question 3

Given $\overrightarrow{OP}=\underset{\sim}{i}+3\underset{\sim}{j}-7\underset{\sim}{k}$ and $\overrightarrow{OQ}=5\underset{\sim}{i}-2\underset{\sim}{j}+4\underset{\sim}{k}$.

i. Find the unit vector in the direction of $\overrightarrow{PQ}$.

OP = Matrix([1, 3, -7])
OQ = Matrix([5, -2, 4])
PQ = OQ - OP
display(PQ)                                 # <4, -5, 11>
display(sqrt(PQ.dot(PQ)))                   # 9*sqrt(2)
display(PQ / sqrt(PQ.dot(PQ)))              # (1/(9*sqrt(2)))<4,-5,11>

ii. Find the direction cosines of $\overrightarrow{PQ}$.

unit_PQ = PQ / sqrt(PQ.dot(PQ))
display(list(unit_PQ))     # [cos(alpha), cos(beta), cos(gamma)]

So $\cos\alpha=\frac{4}{9\sqrt{2}}$, $\cos\beta=\frac{-5}{9\sqrt{2}}$ and $\cos\gamma=\frac{11}{9\sqrt{2}}$.

iii. Find the vector of magnitude $5$ in the direction of $\overrightarrow{QP}$, in polar form.

display(5 * (-unit_PQ))                          # the vector itself
angles = [acos(e) * 180/pi for e in (-unit_PQ)]  # polar angles of QP
display([sy.N(a, 6) for a in angles])            # 108.32, 66.87, 149.80 degrees

The vector of magnitude $5$ in the direction of $\overrightarrow{QP}$ is $5\langle\cos 108.32^\circ,\cos 66.87^\circ,\cos 149.80^\circ\rangle$.

Question 4

$A(1,2)$ and $B(3,4)$.

i. Vector equation of the line $L$ through $A$ and $B$.

A = Matrix([1, 2])
B = Matrix([3, 4])
AB = B - A
display(AB)                # <2, 2>

So $L:\underset{\sim}{r}=\langle 1,2\rangle+t\langle 2,2\rangle$.

ii. Points on the line for $t=0,1,\ldots,5$.

display([(ti, A + ti*AB) for ti in range(6)])

$t$	$\underset{\sim}{r}$
0	$\langle 1,2\rangle$
1	$\langle 3,4\rangle$
2	$\langle 5,6\rangle$
3	$\langle 7,8\rangle$
4	$\langle 9,10\rangle$
5	$\langle 11,12\rangle$

Question 5

A unit vector $\vec{a}$ makes an angle of $\pi/3$ with $i$, $\pi/4$ with $j$ and an acute angle $\theta$ with $k$. Find $\theta$ and the components of $\vec{a}$.

The components of a unit vector are exactly its direction cosines, and $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$.

cos_a = cos(pi/3)
cos_b = cos(pi/4)
cos2_g = 1 - cos_a**2 - cos_b**2
display(simplify(cos2_g))                     # 1/4

theta = acos(sqrt(cos2_g))                    # acute angle only
display(theta, sy.N(theta*180/pi, 6))         # pi/3, 60 degrees

a = Matrix([cos_a, cos_b, cos(theta)])
display(simplify(a), sqrt(a.dot(a)))          # <1/2, 1/sqrt(2), 1/2>, |a| = 1

Hence $\theta=60^\circ$ and $\vec{a}=\frac{1}{2}i+\frac{1}{\sqrt{2}}j+\frac{1}{2}k$.

note

The printed solution also lists the $120^\circ$ case $\vec{a}=\frac{1}{2}i+\frac{1}{\sqrt{2}}j-\frac{1}{2}k$. Because the question specifies an acute angle, that case must be discarded — taking acos of the positive root gives the acute angle directly.

Question 6

If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=8$, find $|\vec{x}|$.

Expanding the dot product gives $|\vec{x}|^{2}-|\vec{a}|^{2}=8$, and $|\vec{a}|=1$.

X = sy.Symbol('X', nonnegative=True)          # X = |x|
display(solve(sy.Eq(X**2 - 1, 8), X))         # [3]

So $|\vec{x}|=3$ (the negative root is rejected because a magnitude cannot be negative).

Question 7

Find the gradient of $f=\dfrac{2x^{2}+y}{x^{2}-y^{2}}$.

f = (2*x**2 + y)/(x**2 - y**2)
grad_f = gradient(f, (x, y))
display(simplify(grad_f[0]))                  # -2*x*y*(2*y + 1)/(x**2 - y**2)**2
display(simplify(grad_f[1]))                  # (x**2 + 4*x**2*y + y**2)/(x**2 - y**2)**2

$\nabla f=\frac{2xy(-2y-1)}{(x^{2}-y^{2})^{2}}\,\boldsymbol{i}+\frac{x^{2}+4x^{2}y+y^{2}}{(x^{2}-y^{2})^{2}}\,\boldsymbol{j}$

Question 8

Calculate the divergence of the vector fields.

a. $F=y^{3}\boldsymbol{i}+xy\boldsymbol{j}$

display(divergence(Matrix([y**3, x*y]), (x, y)))       # x

b. $G=\dfrac{4y}{x^{2}}\boldsymbol{i}+\sin(y)\boldsymbol{j}+3\boldsymbol{k}$

display(simplify(divergence(Matrix([4*y/x**2, sin(y), 3]))))   # cos(y) - 8*y/x**3

c. $G=e^{x}\boldsymbol{i}+\ln(xy)\boldsymbol{j}+e^{xyz}\boldsymbol{k}$

display(simplify(divergence(Matrix([exp(x), ln(x*y), exp(x*y*z)]))))   # e**x + 1/y + x*y*e**(x*y*z)

Question 9

Calculate the curl of the vector fields.

a. $F=3x^{2}\boldsymbol{i}+2z\boldsymbol{j}-x\boldsymbol{k}$

display(curl(Matrix([3*x**2, 2*z, -x])))               # <-2, 1, 0>

b. $F=y^{3}\boldsymbol{i}+xy\boldsymbol{j}-z\boldsymbol{k}$

display(curl(Matrix([y**3, x*y, -z])))                 # <0, 0, y - 3*y**2>

The curl points along $\boldsymbol{k}$ only.

c. $F=(1+y+z^{2})\boldsymbol{i}+e^{xyz}\boldsymbol{j}-(xyz)\boldsymbol{k}$

display(simplify(curl(Matrix([1 + y + z**2, exp(x*y*z), -x*y*z]))))

$\nabla\times F=-(xz+xy\,e^{xyz})\,\boldsymbol{i}+(2z+yz)\,\boldsymbol{j}+(yz\,e^{xyz}-1)\,\boldsymbol{k}$

note

The printed solution drops the minus sign on $F_{3}=-xyz$ in the $\boldsymbol{i}$ and $\boldsymbol{j}$ components and mislabels the $\boldsymbol{j}$ term as $\boldsymbol{k}$. SymPy differentiates each component literally, so the signs come out right.

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Extra Learning Resources

Key Concepts to Master

1. Magnitude: $|\vec{a}|=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}$
2. Unit vector: $\hat{a}=\vec{a}/|\vec{a}|$
3. Direction cosines: the components of a unit vector, with $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$
4. Cross product: $\vec{a}\times\vec{b}$ is perpendicular to both, with $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$
5. Del operator: $\nabla f$ (gradient, a vector), $\nabla\cdot F$ (divergence, a scalar), $\nabla\times F$ (curl, a vector)

SymPy Resources

• SymPy Matrices — dot, cross, norm, jacobian
• SymPy Calculus — the diff used by the helpers above
• SymPy Vector module — CoordSys3D, gradient, divergence, curl as ready-made functions

Common Pitfalls

• Matrix([1,2,3]) is a column vector; dot and cross only work on Matrix objects, not plain lists
• Use ** for powers, and remember sec(x) = 1/cos(x)
• solve returns a list — index it ([0]) or filter it before using a root
• Always declare an applied variable as positive (sy.Symbol('r', positive=True)) to discard non-physical roots