Tutorial 4

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Parallel vectors and the cross product
• Shortest distance between skew lines
• Distance from a point to a line
• Vector, parametric and Cartesian equations of lines
• Equations of planes

Introduction

Vectors are Matrix objects. The cross product is A.cross(B), the dot product is A.dot(B), and A.norm() gives the magnitude $|\vec{a}|$.

import sympy as sy
from sympy.abc import x, y, z, t
from sympy.solvers import solve
from sympy import sqrt, sin, cos, pi, Matrix, simplify, nsimplify
sy.init_printing(use_latex=True)

Question 1

Pipe A has diameter $0.8$ and axis through $(2,5,3)$ and $(7,10,8)$. Pipe B has diameter $1.0$ and axis through $(0,6,3)$ and $(-12,0,9)$. Do the pipes intersect?

The axes are skew lines, so the shortest distance between them is $\left|(\mathbf{a}-\mathbf{b})\cdot\dfrac{\mathbf{n}}{|\mathbf{n}|}\right|$ where $\mathbf{n}=\mathbf{d}_{A}\times\mathbf{d}_{B}$.

a, dA = Matrix([2, 5, 3]),  Matrix([5, 5, 5])       # point and direction of pipe A
b, dB = Matrix([0, 6, 3]),  Matrix([-12, -6, 6])    # point and direction of pipe B

n = dA.cross(dB)
display(n, n/30)                                    # 30*<2,-3,1>, i.e. <2,-3,1>

dist = abs((a - b).dot(n)) / sqrt(n.dot(n))
display(dist, sy.N(dist, 4))                        # sqrt(14)/2 = 1.871

The sum of the radii is $0.4+0.5=0.9\ \text{m}$, which is less than $1.871\ \text{m}$, so the pipes do not intersect and no realignment is required.

Question 2

Determine whether each pair of vectors is parallel.

a1, b1 = Matrix([2, 4, -1]),  Matrix([-6, -12, 3])
display(b1 == -3*a1)                 # True  -> b1 = -3*a1, so parallel

a2, b2 = Matrix([4, 10]),     Matrix([2, -9])
display(sy.simplify(a2[0]/b2[0]), sy.simplify(a2[1]/b2[1]))   # 2 and -10/9 -> not equal

i. $\underset{\sim}{b}=-3\underset{\sim}{a}$, so the vectors are parallel.

ii. The two ratios $4/2=2$ and $10/(-9)=-10/9$ differ, so no single scalar works and the vectors are not parallel.

Question 3

Find the unit vectors perpendicular to $\underset{\sim}{a}$ and $\underset{\sim}{b}$.

i. $\underset{\sim}{a}=\langle 2,4,5\rangle$, $\underset{\sim}{b}=\langle 1,2,-2\rangle$

a = Matrix([2, 4, 5]); b = Matrix([1, 2, -2])
c = a.cross(b)
display(c)                                    # <-18, 9, 0>
display(c.dot(a), c.dot(b))                   # 0, 0  -> perpendicular to both

unit = c / sqrt(c.dot(c))
display(unit)                                 # <-2/sqrt(5), 1/sqrt(5), 0>
display(-unit)                                # the opposite direction is also valid

The two unit vectors are $\pm\left\langle-\dfrac{2}{\sqrt{5}},\dfrac{1}{\sqrt{5}},0\right\rangle$.

ii. $\underset{\sim}{a}=\langle 2,4,-4\rangle$, $\underset{\sim}{b}=\langle 1,2,-2\rangle$

a = Matrix([2, 4, -4]); b = Matrix([1, 2, -2])
display(a.cross(b))                           # <0, 0, 0> -> a = 2*b, they are parallel

The cross product is the zero vector, so $\underset{\sim}{a}$ and $\underset{\sim}{b}$ are parallel. A unit vector perpendicular to both does not exist (it would require dividing by $|\mathbf{0}|=0$).

note

The printed solution computes $a\times b$ as $\langle-2,-1,0\rangle$ with the unit vector $\langle-2/\sqrt{5},-1/\sqrt{5},0\rangle$. That vector is not even perpendicular to $\underset{\sim}{a}$: $\langle-2,-1,0\rangle\cdot\langle2,4,5\rangle=-8\neq0$. The sign error comes from writing $i(-8+10)$ instead of $i(-8-10)$. SymPy gives the correct $a\times b=\langle-18,9,0\rangle$.

Question 4

Find the distance from $P(-3,7,4)$ to the line $r=(2,-2,-3)+\lambda(4,-5,3)$.

The distance is $\dfrac{|\mathbf{u}\times\mathbf{v}|}{|\mathbf{u}|}$ where $\mathbf{u}=(4,-5,3)$ and $\mathbf{v}=\overrightarrow{AP}=\mathbf{p}-\mathbf{a}$.

p = Matrix([-3, 7, 4])
a = Matrix([2, -2, -3])            # point on the line
u = Matrix([4, -5, 3])             # direction of the line
v = p - a
display(v)                         # <-5, 9, 7>

cr = u.cross(v)
display(cr)                        # <-62, -43, 11>
display(sqrt(cr.dot(cr)), sqrt(u.dot(u)))      # 3*sqrt(646), 5*sqrt(2)

display(simplify(sqrt(cr.dot(cr)) / sqrt(u.dot(u))))   # 3*sqrt(323)/5

So the distance is $\dfrac{3\sqrt{323}}{5}\approx10.78$.

Question 5

Distance between the skew lines $l:r=\mathbf{a}+\lambda\mathbf{u}$ and $m:r=\mathbf{b}+\mu\mathbf{v}$.

a = Matrix([0, 4, -1]);  u = Matrix([1, -3, -2])
b = Matrix([2, -1, 0]);  v = Matrix([-3, 1, 2])

display(b - a)                     # <2, -5, 1>
cr = u.cross(v)
display(cr, sqrt(cr.dot(cr)))      # <-4, 4, -8>, 4*sqrt(6)
display(abs((b - a).dot(cr)) / sqrt(cr.dot(cr)))     # 3*sqrt(6)/2

The distance is $\dfrac{9}{\sqrt{6}}=\dfrac{3\sqrt{6}}{2}\approx3.674$.

Question 6

Line $L$ through $P(3,1,-2)$ and $Q(-2,7,-4)$.

i. Vector equation

P = Matrix([3, 1, -2]); Q = Matrix([-2, 7, -4])
PQ = Q - P
display(PQ)                        # <-5, 6, -2>

$L:\underset{\sim}{r}=\langle 3,1,-2\rangle+t\langle-5,6,-2\rangle$.

ii. Parametric equation

display(P + t*PQ)                  # <3 - 5*t, 1 + 6*t, -2 - 2*t>

$x=3-5t,\quad y=1+6t,\quad z=-2-2t,\qquad t\in\mathbb{R}$.

iii. Cartesian equation

Eliminating the parameter $t$ gives $\dfrac{x-3}{-5}=\dfrac{y-1}{6}=\dfrac{z+2}{-2}$. A quick check: every point on the line must make all three ratios equal.

pt = P + 1*PQ                              # t = 1 gives the point Q
display(pt)                                # <-2, 7, -4>
display([sy.simplify((pt[i] - P[i]) / PQ[i]) for i in range(3)])   # [1, 1, 1]

Question 7

Cartesian equation of the plane through $(1,2,-1)$ perpendicular to the line of intersection of $2x+y+z=2$ and $x+2y+z=3$.

n1 = Matrix([2, 1, 1]); n2 = Matrix([1, 2, 1])
n = n1.cross(n2)
display(n)                                    # <-1, -1, 3>

k = n.dot(Matrix([1, 2, -1]))
display(k)                                    # -6

The plane is $-x-y+3z=-6$, i.e. $x+y-3z=6$.

Question 8

Plane containing $L_{1}:\underset{\sim}{r}=\langle 1,-3,4\rangle+t\langle 2,1,1\rangle$ and parallel to $L_{2}:\underset{\sim}{r}=\langle 0,0,0\rangle+s\langle 1,2,3\rangle$.

u = Matrix([2, 1, 1]); v = Matrix([1, 2, 3])
n = u.cross(v)
display(n)                                    # <1, -5, 3>
display(n.dot(Matrix([1, -3, 4])))            # 28
display(n.dot(v))                             # 0 -> v lies in the plane

So the plane is $x-5y+3z=28$. Because $\mathbf{n}\cdot\mathbf{v}=0$, the normal is perpendicular to the direction of $L_{2}$, which proves $L_{2}$ is parallel to the plane.

Question 9

Plane containing $L_{1}:\underset{\sim}{r}=\langle-2,3,4\rangle+t\langle 1,2,-1\rangle$ and $L_{2}:\underset{\sim}{r}=\langle 3,4,0\rangle+s\langle-1,-2,1\rangle$.

A = Matrix([-2, 3, 4]); B = Matrix([3, 4, 0])
AB = B - A
display(AB)                                   # <5, 1, -4>

u = Matrix([1, 2, -1])
n = AB.cross(u)
display(n)                                    # <7, 1, 9>

display(n.dot(A), n.dot(B))                   # 25, 25 -> both points lie on the plane

So the plane is $7x+y+9z=25$. (Using the other point gives the same $k$, which is a quick consistency check.)

Question 10

$\underset{\sim}{a}=\langle 1,-2,-3\rangle$, $\underset{\sim}{b}=\langle 2,1,-1\rangle$, $\underset{\sim}{c}=\langle 1,3,-2\rangle$.

i. $\underset{\sim}{a}\cdot\underset{\sim}{b}\left(\underset{\sim}{a}\times\underset{\sim}{b}\right)$

a = Matrix([1, -2, -3]); b = Matrix([2, 1, -1]); c = Matrix([1, 3, -2])
display(a.dot(b))                             # 3
display(a.cross(b))                           # <5, -5, 5>
display(a.dot(b) * a.cross(b))                # <15, -15, 15>

ii. $(\underset{\sim}{a}+\underset{\sim}{b})\times\underset{\sim}{c}$

display(a + b)                                # <3, -1, -4>
display((a + b).cross(c))                     # <14, 2, 10>

Question 11

Shortest distance between the skew lines $L_{a}:r_{a}=4i+2j-6k+t(2i-j-k)$ and $L_{b}:r_{b}=i-3j-3k+t(i-2j-k)$.

ba = Matrix([2, -1, -1]); bb = Matrix([1, -2, -1])
aa = Matrix([4, 2, -6]);  ab = Matrix([1, -3, -3])

cr = ba.cross(bb)
display(cr, sqrt(cr.dot(cr)))                 # <-1, 1, -3>, sqrt(11)

display(abs((aa - ab).dot(cr)) / sqrt(cr.dot(cr)))    # sqrt(11)

The shortest distance is $\dfrac{11}{\sqrt{11}}=\sqrt{11}\approx3.317$.

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Extra Learning Resources

Key Concepts to Master

1. Parallel test: $\vec{a}\times\vec{b}=\mathbf{0}$
2. Perpendicular test: $\vec{a}\cdot\vec{b}=0$
3. Distance point to line: $\dfrac{|\mathbf{u}\times\mathbf{v}|}{|\mathbf{u}|}$
4. Distance between skew lines: $\left|(\mathbf{b}-\mathbf{a})\cdot\dfrac{\mathbf{u}\times\mathbf{v}}{|\mathbf{u}\times\mathbf{v}|}\right|$
5. Plane from a point and normal: $\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_{0})=0$

SymPy Resources

• SymPy Matrices — dot, cross, norm, det
• SymPy solve — solving the simultaneous equations that define a plane

Common Pitfalls

• Cross product is not commutative: a.cross(b) == -b.cross(a)
• Matrix is a column vector, so Matrix([2,5,3]) is a point in 3D, not a row
• The perpendicular vector from a cross product can be scaled by any non-zero number — always divide by the magnitude to get the unit vector
• When a plane is given by three points, substitute a point into $k$ only after computing the normal