Defining a function \(f,\)
the notation \(f(x)\) is read as “\(f\) of \(x\)”
and represents the output that results
from evaluating \(f\) at the input \(x,\)
generally expressed as a formula.
The functions most often used in this course
will be those conventionally called trigonometric
and inverse trigonometric functions.
The notation \(\mathtt{trig}\) will be used to represent
a generic trigonometric function
— any of \(\sin\) or \(\cos\) or \(\tan\)
or \(\csc\) or \(\sec\) or \(\cot\) —
and instead of \(\mathtt{trig}^{-1}\) which is typical in the US
we’ll use \(\mathrm{arc}\mathtt{trig}\) to represent
a generic “inverse” trigonometric function.
Characters typeset in a roman font like “\(\mathrm{e}\)” or “\(\mathbf{j}\)”
or “\(\operatorname{proj}\)” represent specific constants or named functions,
whereas italicized characters like “\(x\)” or “\(f\)” are indeterminate variables or functions.
Bold characters like \(\bm{v}\) are vector-valued,
whereas non-bold characters like \(v\) will be numerically valued.
Variables representing angles will be denoted by lowercase Greek letters,
usually \(\vartheta\) and \(\varphi,\)
or sometimes \(\alpha\) and \(\beta\) and \(\gamma.\)
Rectangular Geometry Fundamentals
work in slopes of perpendicular lines
and how to calculate the distance to a line from a point.
Draw the coordinate axes.
Show them the rectangular grid.
Label the quadrants.
Plot the point \((2,-3)\)Sketch a random point and use a meter stick to determine its coordinates.
The defining feature of the “shape” of this plane
is how you define distance.
Note: this is a choice we’re making.
Distance formula \(\operatorname{d}(A,B)\):
it’s just another perspective on the Pythagorean theorem.
(Pythagoras was a cult leader btw)
(Later at the end of class I’ll prove the Pythagorean theorem)
The midpoint formula \((\overline{x}, \overline{y})\):
the coordinates of the midpoint
are just the averages of the coordinates of the endpoints.
Calculate distance between \((4,8)\) and \((2,-3).\)Calculate midpoint between \((4,8)\) and \((2,-3).\)
Slight pivot:
recall that any equation containing \(x\) and \(y\)
corresponds to a curve in the \(xy\)-plane
consisting of all points \((x,y)\)
that satisfy the equation.
EXAMPLES FROM COLLEGE ALGEBRA REAL QUICK.
Let’s talk about a circle though
a curve that is defined as all points
that are a fixed distance (radius) from a given point (center).
The equation corresponding to a circle
is just the distance formula with one of the points
kept arbitrary \((x,y).\)
What’s the distance between \(4,8)\) and \((x,y)?\)
What’s an equation that represents all points \((x,y)\)
such that the distance between \(4,8)\) and \((x,y)\) is 7?
Facts: area and circumference of a circle.
Note \(\pi \approx 3.1415926535897932384626433832795028841968.\)
Prove distance formula / Pythagorean theorem.
The slope of the line containing those two points is
\[ m = \frac{\text{rise}}{\text{run}} = \frac{y_2-y_1}{x_2-x_1}; \]
this is the “rate” at which the line is increasing/decreasing.
It's also referred to as the pitch or grade.
The “point-slope” form, “slope-intercept” form,
and “general” form of a line are respectively are
\[ (y-y_1) = m(x-x_1) \qquad y = mx+b \qquad Ax + By = C \,, \]
but these names are stupid.
What’s an equation for the line with \(y\)-intercept 5 and slope \(\frac{2}{3}?\) What’s an equation for the line with slope \(\frac{2}{3}\) passing through the point \((4,8)?\) What’s an equation for the line passing through the points \((4,8)\) and \((2,-3)?\)
Horizontal and vertical lines
have special forms, \(y=b\) and \(x=a\) respectively.
Two lines are parallel
if they never intersect, and so have the same slope.
Two lines are perpendicular (or normal, or orthogonal)
if they intersect at a right angle,
so one’s slope will be the negative reciprocal of the other.
What’s an equation for the line parallel to \(y = -5x + 3\) that passes through the point \((2,-3)?\) What’s an equation for the line perpendicular to \(y = -5x + 3\) that passes through the point \((2,-3)?\)
The unit circle has radius 1 and passes through the origin.
What’s an equation for the unit circle? Where does the unit circle intersect the line \(y = \sqrt{3}x?\)
Angle Measure & Technology Demo
Bring a straightedge and giant protractor for demonstration.
“sexigesimal”
Draw and define angles, initial side and terminal side.
Positive angles are clockwise — right-hand rule.
Coterminal angles and principal angles.
Same as an arclength having a “reference number”.
The “co” in coterminal is the same one as is cosine and cotangent and cosecant.
Radian measure: aligns with the “unit circle” approach,
and lends itself to geometry.
Draw a picture of the unit circle as the angle
of a sector of a larger circle.
For an angle of radian measure \(\theta\)
within a sector with initial and terminal side lengths \(r,\)
the length of the arc subtending the angle will be \(r\theta\)
and the area of the sector will be \(\frac{1}{2}r^2\theta.\)
Degree measure: more often used in some fields
(y’all’s fields I think) so from now on out we’ll use it here.
Based on Babylonian base-60 math, and ~365 days in a year.
Cut a full rotation into 360 “pieces” each called a degree (1°).
Always put the degree sign.
So for a full rotation, \(360° = 2\pi\) in radians.
Half a rotation, \(180° = \pi\) in radians.
A right angle is 90°. A “diagonal” angle is 45°.
Draw an angle on the whiteboard
and literally measure it with the giant protractor.
How many degrees is \(\pi/3\) radians?
How many degrees is \(\pi/4\) radians?
How many degrees is \(\pi/6\) radians?
How many degrees is 1 radian?
How many radians is 77°?
What is the degree and radian measure
of the principal angle coterminal to 1776°?
Fractional degrees: either a decimal number of degrees (DD)
or degrees/minutes/seconds (DMS),
which is more often used in terrestrial navigation, GPS.
Latin “pars minuta prima” = “first small part”
and “pars minuta secunda” = “second small part”
for minutes and seconds respectively.
Whereas “radian” was coined in 1871
How many degrees is 33°22′11″ expressed as a decimal?
Polar Geometry Fundamentals
Bring giant protractor and class protractors
for demonstration and exercises.
Fundamental Question:
Given a point in space,
how do we describe its location?
How do we communicate where that point is?
Rectangular coordinate review before introducing polar coordinates:
pole and polar axis.
Right-hand rule again.
Pull up Desmos to show the different grid systems.
Draw coordinate axes and a point on the whiteboard
and use the giant protractor
to literally measure the polar coordinates of that point.
Carefully demonstrate how to convert between
rectangular and polar coordinates.
Sine and cosine are percentages.
Arctangent, an “inverse” of tangent,
takes a slope and returns and angle.
It’s the tan-1 button on your calculator.
“co”sine is literally the angle compliment of sine
What are the polar coordinates of the point
with rectangular coordinates \((1,2)?\)
What are the rectangular coordinates of the point
with polar coordinates \((6,27°)?\)
What are the polar coordinates of the point
with rectangular coordinates \((-7,3)?\)
Mention atan2(y,x).
For a given point there are infinitely many pairs
of polar coordinates that describe that point.
E.g. (1,23°) and (1,383°) and (1,-337°) and (-1,203°)
all describe the same point,
but (1,23°) is “principal” among them since it’s simplest in a sense.
For a given point only one pair of polar coordinates \((r,\theta)\)
with \(r \geq 0\) and \(0 \leq \theta \lt 360°\) that describes it.
What are the “principal” polar coordinates of the point
with point coordinates \((3,1776°)?\)
Terrestrial navigation relies heavily
on rectangular/polar coordinates.
Draw a
horizontal coordinate system
and label the
observer,
horizon (circle around observer),
zenith (straight above the observer, normal to the earth),
some object in the sky,
its azimuth
and its altitude (or altitude angle),
and use the words
elevation (or inclination)
and depression (or declination).
Draw a plane from a top-down view and label its
heading (direction it’s pointing/facing)
and its track (the direction its actually going)
and its course (intended direction)
which might be different because of the wind or something.
Draw some other object and describe its
bearing relative to the plane and relative to north.
Directions are usually described as an angle
measured clockwise from north (or south).
E.g. “N 34°56′ W” refers to the direction
34°56′ rotated counterclockwise from north.
Whether “north” refers to geographic (true) north
or magnetic north or grid north depends on context.
If time permits describe
lines of latitude
and lines of longitude
and the prime meridian and its history
and GPS coordinates on earth.
The radius of the earth is about 3960 miles.
“Unit Circle” Trigonometry
The unit circle is the circle of radius one centered at the origin.
Its circumference is \(2\pi.\)
Show that the point
\(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)
is on the unit circle.
Helpful to find the coordinates of the terminal point of \(\pi/6\) later.
Starting at the point \((1,0)\)
and moving around the circle counterclockwise by an arclength of \(t\)
we land at the terminal point corresponding to \(t.\)
Negative values of \(t\) correspond to clockwise movement.
What “terminal point” corresponds to
\(\pi/2?\) or \(-\pi?\) or \(7\pi?\)
Have a running “unit circle” up on the white board.
Sometimes finding the coordinates of the terminal point is easy
but there are other \(t\) for which this question
requires clever geometry to answer.
What “terminal point” corresponds to
\(\pi/4?\) or \(\pi/6?\) or \(\pi/3?\)
Note that these are just nice fractions of a rotation.
Gotta refer to simply symmetry for \(\pi/4.\)
Note that \(1/\sqrt{2} = \sqrt{2}/2.\)
The coordinate for \(\pi/6\) are actually a bit tricky:
note that the terminal point \((x,y)\) for \(\pi/6\)
has the property that it’s equidistant to
\((0,1)\) and to \((x,-y),\)
which means \(2y = \sqrt{(x-0)^2+(y-1)^2}.\)
Solve this equation simultaneously with \(x^2+y^2=1.\)
Then argue \(\pi/3\) from symmetry.
Many different \(t\) correspond to the same point,
which is a bit silly. Let’s be more pragmatic.
Given a real number \(t\) its reference number \(\overline{t}\)
is the number between \(0\) and \(2\pi\) that has the same terminal point as \(t.\)
What is the “reference number” of
\(15\pi?\) or \(-13\pi/3?\) or \(7?\)
Sometimes finding the coordinates of the terminal point
for a value of \(t\) can’t be done with even the most clever geometry.
What “terminal point” corresponds to \(t=1?\)
Let’s at least get some vocab:
the terminal point for \(t\) has coordinates
\(\bigl(\cos(t), \sin(t)\bigr)\).
To rephrase that last question,
What are the values of \(\cos(1)\) and \(\sin(1)?\)
Make sure calculators are in radian mode!
That’s really all the trig functions sine and cosine are:
functions that take an arclength along the unit circle
and return the coordinates of the corresponding terminal point.
Later we’ll talk about other interpretations of sine and cosines.
Quick etymological aside:
The word has come with some distortion
from Sanskrit through Arabic and Latin.
Accounts differ on the details but the basic story is this:
the Sanskrit jya ("chord") was taken into Arabic as jiba
but the word that was translated into Latin was not this word but jaib ("bay")
and this became sinus ("bay" or "curve") which was anglicized as sine.
What is the value of
\(\cos(\pi/4)?\) or \(\sin(\pi/6?)\) or \(\sin(\pi/3?)\)
or \(\cos(15\pi)?\) or \(\sin(13\pi/3)?\)
\(\cos(-\pi/4)?\) or \(\sin(5\pi/6?)\) or \(\sin(-7\pi/3?)\)
Talk about positivity and negativity of sine/cosine.
What about for negative arguments?
Note \(\cos(-t) = \cos(t)\) and \(\sin(-t) = -\sin(t).\)
Furthermore, you really just gotta know about the first quadrant,
and translate everything from there as needed.
Summarize everything with the classic “unit circle” drawing.
Typographic note:
some authors write their trig functions without parenthesis;
know that \(\sin t\) and \(\sin(t)\) are the same.
Also it’s conventional to write \(\sin^2(t)\)
for \(\bigl(\sin(t)\bigr)^2.\)
However \(\sin^{-1}(t) \neq \bigl(\sin(t)\bigr)^{-1};\)
a power of \(-1\) on a function denotes its inverse function.
Because they’re the \(x\)- and \(y\)-coordinate
on a circle of radius one, \(\sin^2(t) + \cos^2(t) = 1.\)
This is called a Pythagorean identity.
There are four other trig functions to know:
tangent and secant
and cotangent and cosecant.
Tangent is the only special one:
it represents the slope of the line
passing through the origin and the terminal point of \(t.\)
The others are just a notational convenience.
What is the value of
\(\tan(\pi/4)?\) or \(\tan(\pi/6?)\) or \(\cot(\pi/6?)\)
or \(\sec(\pi/4)?\) or \(\csc(-\pi/3?)\) or \(\cot(\pi/6?)\)
\(\csc(-\pi/4)?\) or \(\sec(7\pi/6?)\) or \(\tan(-7\pi/3?)\)
There are two other Pythagorean identities
based on these other trig functions:
\(\tan^2(t) + 1 = \sec^2(t)\) and \(1 + \cot^2(t) = \csc^2(t).\)
Goals:
commit all this to memory
and
review graphs and “transformations of functions”.
What is the value of \(\mathtt{trig}(\pm n\pi/k)?\)
To craft general exercises of this form,
roll a d6 for the \(\mathtt{trig}\) function,
a d8 for \(\pm k,\) and a d20 literally for \(n.\)
\(1: \sin\)
\(2: \cos\)
\(3: \tan\)
\(4: \csc\)
\(5: \sec\)
\(6: \cot\)
\(1: -6\)
\(2: -4\)
\(3: -3\)
\(4: -2\)
\(5: 2\)
\(6: 3\)
\(7: 4\)
\(8: 6\)
Planar Terrestrial Navigation
"Geometry" literally "earth measure".
Gnomon used to measure time of day.
Trying to calculate earth's size.
26,700 year wobble of the poles.
procession of equinoxes.
"science of artillery" inspiring invention of trig.
also invention of pendulum and time keepers.
History of sine and cosine;
Richard Norwood formalized the notation we used today in the 1500s–1600s.
Map making and projections: conformal maps.
A path of constant bearing
is a rhumb line is a loxodrome,
which is not a great circle, a orthodrome.
Secants of latitude and Mercator’s projection.
No map of the earth preserves all three of distance, shape, and direction
(which maps preserve 2/3 though?)
At what size of measure to surveyors start to account for the earths curvature.
“Right Triangle” Trigonometry
In addition to coordinates on the unit circle
and in addition to scaling percentages of projections,
sine and cosine and tangent
can be thought of as ratios of the side-lengths
of a right triangle relative to an acute angle \(\theta.\)
(This is the way it’s classically taught in primary school.)
\[
\sin\bigl(\theta\bigr) \!=\! \frac{\text{“opposite”}}{\text{“hypotenuse”}}
\quad \cos\bigl(\theta\bigr) \!=\! \frac{\text{“adjacent”}}{\text{“hypotenuse”}}
\quad \tan\bigl(\theta\bigr) \!=\! \frac{\text{“opposite”}}{\text{“adjacent”}}
\]
Show that its the same as the “projections” interpretation.
The other trigonometric functions
cosecant and secant and cotangent
are still the reciprocals of the original three.
\[
\csc\bigl(\theta\bigr) \!=\! \frac{\text{“hypotenuse”}}{\text{“opposite”}}
\quad\sec\bigl(\theta\bigr) \!=\! \frac{\text{“hypotenuse”}}{\text{“adjacent”}}
\quad\cot\bigl(\theta\bigr) \!=\! \frac{\text{“adjacent”}}{\text{“opposite”}}
\]
And the three Pythagorean identities still hold,
\[
\sin^2\bigl(\theta\bigr) + \cos^2\bigl(\theta\bigr) = 1
\quad\tan^2\bigl(\theta\bigr) + 1 = \sec^2\bigl(\theta\bigr)
\quad1 + \cot^2\bigl(\theta\bigr) = \csc^2\bigl(\theta\bigr)
\]
Those “nice” output values of sine and cosine we’ve committed to memory
correspond to 30°-60°-90° and 45°-45°-90° triangles
where the side-lengths are in a nicely proportional.
Just start drawing right triangles
with a single acute angle and single side-length labelled —
of all different sizes and shapes and orientations
— and fill in the missing side-lengths.
Recall that the angles in any triangle must sum to 180°.
Before proving this let’s introduce a couple vocab terms:
angle compliment (\(90°-\theta\))
and angle supplement (\(180°-\theta\)).
Remember that the word "compliment" here is the samesake of the "co"-
in half of the trigonometric functions.
Draw a generic triangle with angles labelled \(\alpha\) and \(\beta\) and \(\gamma\),
and sketch the line parallel to the base through the opposite vertex
and start labelling complimentary and supplementary angles.
So another way to say that
angles in any triangle must sum to 180°
is to say the pair of acute angles in a right triangles
are complimentary angles.
In all the previous examples, fill in the missing angles.
One last fact: the area of a triangle is ½×(base)×(height)
where “height” is the height of the triangle’s altitude
perpendicular to a side chosen as the “base”.
This is because a triangle always occupies half the space
of the base×height rectangle that “bounds” it.
If we don’t know this height for a given base though,
but we do know the adjacent angle,
we can calculate the area to be ½×(base)×(slant)×\(\sin(\theta).\)
Generically \(\frac{1}{2}AB\sin(\theta)\) as a formula.
Inverse Trigonometric (Cyclometric) Functions
The inverse trigonometric (or cyclometric or arcus) functions —
arccosine, arcsine, arctangent,
arcsecant, arccosecant, and arccotangent
— are the inverses of the trigonometric functions.
They each take a ratio/scaling-factor as input
and return the corresponding angle/arclength as output.
E.g. since \(\tan(45°) = 1\) we also have
\(\operatorname{arctan}\bigl(1\bigr) = 45°.\)
The notation \(\operatorname{arctan}\) and \(\tan^{-1}\)
mean the same thing, and tan-1
is how most calculators are labelled.
Draw a bunch of examples of right triangles
with only two side-lengths labelled,
and figure out the measure of the unlabelled angles.
There is a cute class of exercises
I always see with these things:
If \(\operatorname{arctan}\Bigl(\frac{5}{2}\Bigr) = \theta\)
and \(\theta\) is an acute angle.
what must \(\sin(\theta)\) be?
If \(\operatorname{arccos}\Bigl(\frac{\sqrt{x}}{x-1}\Bigr) = \theta\)
and \(90° \leq \theta \leq 180°,\)
what must \(\tan(\theta)\) be?
When working only with right-triangles
outside the context of coordinate geometry
we don’t have to consider the domain and range of these functions.
But in the context of coordinate geometry,
there is a quirk to be aware of here.
Just like the function \(f(x) = x^2\) has no true inverse
but instead only its positive and negative branches
have inverses \(\sqrt{x}\) and \(-\sqrt{x},\)
the definition of the inverse trig functions
depends on a branch cut of their domain.
I.e. whereas \(\tan\bigl(\operatorname{arctan}(x)\bigr) = x,\)
\(\operatorname{arctan}\bigl(\tan(\theta)\bigr) \neq \theta\) in general.
Notice how the following questions are different:
What is the value of \(\operatorname{arcsin}\bigl(\frac{1}{2}\bigr)?\)
For what angles \(\theta\) does \(\sin(\theta) = \frac{1}{2}\)
Briefly, arccosine and arcsecant and arccotangent
only return angles in the upper hemisphere,
and arcsine and arccosecant and arctangent
only return angles in the right hemisphere.
This convention exists
so arcsine and arccosine and arctangent and arccotangent
are “continuous” (don’t have breaks) on their domain.
Then the range for arcsecant nearly matches the range of arccosine
and the range for arccosecant nearly matches the range of arcsine.
The Law of Sines & Law of Cosines
Two-day lecture.
On the board, for reference, draw a generic triangle
with side-lengths labelled \(A\) and \(B\) and \(C\)
and opposing angles labelled
\(\alpha\) and \(\beta\) and \(\gamma\) respectively.
Demonstrate problem-solving at the end.
Motivating Question
Given a generic triangle
with three measurements given among its side-lengths and angles,
how do we calculate the other details of the triangle?
½AB\sin(theta) too
Everything we’ve been doing with right triangles
generalizes to general triangles,
and the vehicles of this generalization are
the law of sines and the law of cosines.
First the law of sines:
Given a triangle with side-lengths \(A\) and \(B\) and \(C\)
and opposing angles \(\alpha\) and \(\beta\) and \(\gamma\) respectively,
\[\frac{\sin(\alpha)}{A} = \frac{\sin(\beta)}{B} = \frac{\sin(\gamma)}{C}\,.\]
Given a triangle with acute angles \(\alpha\) and \(\beta\)
and opposing side-lengths \(A\) and \(B\) respectively,
drop a perpendicular from the unlabelled angle to the unlabelled side,
cutting the original triangle into two right triangles,
and name the length of this perpendicular \(X.\)
Now \(X\) is the “opposite side” of two right triangles,
one with angle \(\alpha\) and hypotenuse \(B\)
and another with angle \(\beta\) and hypotenuse \(A.\)
This indicates that \[X = B\sin(\alpha) = A\sin(\beta),\]
but \(X\) is incidental in this equation;
the equality of the other two expressions in the law of sines:
\[\frac{\sin(\alpha)}{A} = \frac{\sin(\beta)}{B} = \frac{\sin(\gamma)}{C}\,.\]
And this proof even extends to obtuse-angled triangles
by, in the illustration,
extending the base and drawing the perpendicular outside the triangle:
for the obtuse angle \(\alpha\) it’s critical to note
that \(\sin(180°-\alpha) = \sin(\alpha).\)
Draw some generic AAS triangles
with at least one pair of an opposing side/angle
and another angle and fill in the missing measurements.
(law of sines)
Draw an SSA triangle with 2′ and 3′ and 27°
where the angle between the 2′ and 3′ sides is clearly obtuse
and fill in the missing measurements.
(law of sines requiring arcsin)
But there’s a subtlety to that last example.
Our answer is correct if we trust the scale of the drawing,
but there is actually another triangle
that has the same initial measurements.
Draw the other SSA triangle with 2′ and 3′ and 27°
where the angle between the other 2′ and 3′ sides is clearly acute
and fill in the missing measurements.
(law of sines requiring arcsin,
where the target angle is 90° plus what arcsine reports)
Now towards the law of cosines:
The so-called projection laws
follow from the same idea as the law of sines,
except instead of looking at the sine of those angle we look at the cosine.
If the once-unlabelled side of that same triangle is now labelled \(C,\)
the perpendicular bisector \(X\) cuts \(C\) into two lengths.
One of those lengths is \(B\cos(\alpha)\) and the other is \(A\cos(\beta),\)
to together they sum to \(C.\)
Writing this altogether
\( \displaystyle A \!=\! B\cos(\gamma) \!+\! C\cos(\beta) \)
\( \displaystyle B \!=\! A\cos(\gamma) \!+\! C\cos(\alpha) \)
\( \displaystyle C \!=\! A\cos(\beta) \!+\! B\cos(\alpha) \)
The law of cosines follows from an algebraic manipulation
of these projection laws.
Within each projection law,
first multiply through by another copy of the isolated side-length:
\[\begin{cases}
A^2 = AB\cos(\gamma) + AC\cos(\beta)
\\B^2 = AB\cos(\gamma) + BC\cos(\alpha)
\\C^2 = AC\cos(\beta) + BC\cos(\alpha)
\end{cases}\]
Then subtract any two equations from the other,
and cancelling terms and rearranging we get:
\[\begin{align*}
C^2 - A^2 - B^2
&= AC\cos(\beta) + BC\cos(\alpha)
- \bigl( AB\cos(\gamma) + AC\cos(\beta) \bigr)
- \bigl( AB\cos(\gamma) + BC\cos(\alpha) \bigr)
\\ C^2 &= A^2 + B^2 - 2AB\cos(\gamma)
\end{align*}\]
Now the law of cosines:
Prove the Pythagorean theorem (distance formula)
and the law of cosines.
\[C^2 = A^2 + B^2 - 2AB\cos(\gamma)\]
Draw some generic SAS triangles
and fill in the missing measurements.
(law of cosines)
Draw a generic SSS triangles
and fill in the missing measurements.
(law of cosines requiring arccos)
The law of cosines quickly leads to another area formula.
Letting \(S = \frac{1}{2}(A+B+C),\)
a number referred to as the semiperimeter of the triangle,
the area of the triangle will be \(\sqrt{S(S-A)(S-B)(S-C)}.\)
This is called Heron’s formula.
Prove Heron’s formula.
First, note that when solving a SSS triangle with the law of cosines,
if the triangle is obtuse you can run into the same issue
as with SSA triangles in the step after the law of cosines,
so be careful there
Second, remember that this all should break into two sections:
(1) the law of sines and subtleties of SSA triangles, then
(2) the projection laws and law of cosines
and SSS triangle and herons formula that follows.
At some point tell the students that if they’re handed information
that doesn’t sincerely correspond to a triangle,
then trying to solve the remaining sides they’ll run into
a computation error from the domain of arcsine or arccosine or something.
The Law of Sines & Law of Cosines Redux
TK
Graphs of Trigonometric Functions
Today we’re taking this class back to College Algebra for a moment.
One of the major themes of that class
was to build a library of functions,
and to understand the geometry of their graphs.
Today we do the same,
adding all the trigonometric functions to our library
and analyzing their graphs.
Pop open Desmos and explain how the graphs of
sine, cosine, and tangent come from the unit circle,
and how the graphs of cosecant, secant, and cotangent
come from the graphs of their reciprocal counterparts.
Recall how you can sketch the graph of
\(a f\bigl(k(x-b)\bigr)+v,\)
just by knowing the graph of \(f\)
and thinking of the parameters
\(a,\) \(b,\) \(k,\) and \(v\)
as linear “transformations” of the base graph of \(f\)?
We can do the same with trigonometric functions.
Pop open Desmos with the graph of base sine
to progressively transform.
For a sinusoidal curve,
its period is the length of its minimal repeating segment,
its amplitude is the highest value it attains,
and its phase shift is it’s horizontal offset.
The graph of the base sine function
has a period of \(2\pi,\) amplitude of \(1,\)
and no phase shift or vertical shift
What does the graph of \(\sin(x)+3\) look like?
What about \(4\sin(x)?\)
What about \(\sin\bigl(x+\frac{\pi}{2}\bigr)?\)
What about \(\sin(2x)?\)
What about \(\sin(-x)?\)
The transformed graph of \[ a\sin\big(k(x-b)\big) +v \]
will have period \(2\pi/k\), amplitude \(|a|\),
phase shift \(b,\) and vertical shift \(v.\)
Manually sketch a graph of the function
\( 2\sin\bigl(x-\frac{\pi}{3}\bigr). \)
Manually sketch a graph of the function
\( \sin\bigl(2x-\frac{2\pi}{3}\bigr). \)
Manually sketch a graph of the function
\( \frac{1}{2}\cos\bigl(x-\frac{\pi}{6}\bigr). \)
Manually sketch a graph of the function
\( \tan\bigl(\frac{1}{3}x\bigr)+2. \)
Manually sketch a graph of the function
\( 3\sec\bigl(x+\frac{4\pi}{3}\bigr). \)
Trigonometric Identities
Two-day lecture; demonstrate problem-solving
We’re going to speedrun §4 because it’s not that important,
but it’s got some interesting mathematics in it,
and I've been asked to talk about it anyways.
First review of some identities we already know.
Cofunction identities,
and the fact that cosine is an even function
and sine is an odd function:
\(\sin(\theta) = \cos\bigl(90°\!-\theta\bigr)\)
\(\cos(\theta) = \sin\bigl(90°\!-\theta\bigr)\)
\(\cos(-\theta) = \cos(\theta)\)
\(\sin(-\theta) = -\sin(\theta)\)
Then all of these extend naturally
to tangent and secant and cosecant and cotangent too.
Pythagorean identities are chill.
\(\sin^2(\theta) + \cos^2(\theta) = 1\)
\(\tan^2(\theta) + 1 = \sec^2(\theta)\)
\(1 + \cot^2(\theta) = \csc^2(\theta)\)
Sum-of-angles formulas:
I don’t have these memorized, especially not the one for tangent,
but I have to draw the picture to derive them every time.
Then the Double-angle formulas
follow from those, letting \(\theta = \phi,\)
and half-angle formulas follow from those
and the first Pythagorean identity.
Question: who cares?
If you’re interested in the exact forms (not just decimal approximations)
of the outputs of trigonometric functions,
the sum-of-angles, and half-angle formulas can help:
What is the exact value,
expressed in terms of radicals if necessary,
of \(\cos\bigl(15°\bigr)?\)
Note \(\cos\bigl(\frac{\pi}{12}\bigr) = \cos\bigl(\frac{\pi}{3} - \frac{\pi}{4}\bigr)
OR \(\cos\bigl(\frac{\pi}{12}\bigr) = \cos\bigl(\frac{\pi/6}{2}\bigr).\).
One of the classic class of exercises
is to prove contrived “identities” involving trigonometric functions
that follow from the fundamental identities we’ve discussed.
Verify that
\(\sin(\theta) + \cos(\theta)\cot(\theta) = \csc(\theta).\)
Fermat primes are those of the form \(2^{2^n}+1.\)
\( \cos(36°) = \frac{\sqrt{5}+1}{4} \)
\[ \cos\Bigl(\frac{\pi}{17}\Bigr) = \frac{15+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}{4\sqrt{2}} \)
\[ \cos\Bigl(\frac{\pi}{257}\Bigr) = \frac{}{} \)
\[ \cos\Bigl(\frac{\pi}{65537}\Bigr) = \frac{}{} \)
Harmonic Motion
TK
Vectors in Two-Dimensional Space
Sketch the coordinate axes for positive \(x\)
and sketch the vectors \(\bm{u} = \langle 3,4\rangle\)
and \(\bm{v} = \langle 5,-2\rangle\)
as running examples to use for all the following definitions.
Vectors help us talk about geometry beyond two-dimensional space.
In particular, we need them to talk about “directions”
since talking about a “slope” is no longer sufficient.
Define vector \(\bm{v} = \langle v_1,v_2 \rangle\)
and its the horizontal and vertical components.
A point and a vector carry the same data, but with a different connotation:
a point is static whereas a vector denotes movement —
like a displacement or velocity or force.
Define initial and terminal point
and the \(\overrightarrow{AB}\) notation,
and the nuance of the initial point being the origin by default.
Talk about \(\bm{v}\) versus \(\vec{v}.\)
Define vector versus scalars.
Sum and scalar multiple of vectors (with picture!)
and the resultant vector
both for sum of vectors and difference of vectors.
\[
\bm{u} = \langle u_1, u_2 \rangle
\quad
\bm{v} = \langle v_1, v_2 \rangle
\quad
\implies
\quad
\bm{u} + \bm{v} = \langle u_1+v_1, u_2+v_2 \rangle
\quad
k\bm{u} = \langle k u_1, k u_2 \rangle.
\]
Define length/magnitude of a vector
and the zero vector
and unit vectors \(\bm{\hat{v}}\) (“vee hat”)
and the unit coordinate vectors
\({\mathbf{i} = \langle 1,0 \rangle}\)
and \({\mathbf{j} = \langle 0,1 \rangle}\)
and that we sometimes write \(\bm{v} = \langle v_1, v_2\rangle\)
as \(v_1\mathbf{i} + v_2\mathbf{j}.\)
A vector’s magnitude encodes not only its direction
but the magnitude in which it is heading in that direction.
If a vector \(\bm{v}\) is an airplane’s velocity vector,
then \(|\bm{v}|\) is the plane’s speed in that direction.
Suppose an airplane has velocity vector
of \(\langle -192, 494 \rangle\) mph;
this is the plane’s heading.
What is the plane’s speed?
How do you express this heading as an angle?
Suppose an airplane has velocity vector
of \(\langle -192, 494 \rangle\) mph,
but there is a 30 mph wind from the south.
What is the plane’s speed and course?
The Dot Product & Projections
The language of vectors gives us a simpler computation
to talk about the angle between them.
Given two vectors
\(\bm{u} = \langle u_1,u_2 \rangle\)
and \(\bm{v} = \langle v_1,v_2 \rangle,\)
their dot product \(\bm{u}\cdot\bm{v}\)
(sometimes called the scalar product or inner product)
is the number defined as the result
of the calculation \(\bm{u}\cdot\bm{v} = u_1v_1+u_2v_2.\)
Sketch the coordinate axes for positive \(x\)
and sketch the vectors \(\bm{u} = \langle 3,4\rangle\)
and \(\bm{v} = \langle 5,-2\rangle\)
and compute \(\bm{u}\cdot\bm{v}.\)
Alternatively if \(\theta\) is the angle between \(\bm{u}\) and \(\bm{v},\)
then \(\bm{u}\cdot\bm{v} = |\bm{u}||\bm{v}|\cos(\theta).\)
These definitions are equivalent due to the law of cosines.
Prove \(\bm{u}\cdot\bm{v} = |\bm{u}||\bm{v}|\cos(\theta).\)
Write the law of cosines as
\({ |\bm{u} - \bm{v}|^2 = |\bm{u}|^2 + |\bm{v}|^2 - 2|\bm{u}||\bm{v}|\cos\(theta)} \)
and expand everything out in terms of components.
Calculate the measure of the angle between
\(\bm{u} = \langle 3,4\rangle\)
and \(\bm{v} = \langle 5,-2\rangle.\)
Note the dot product is commutative,
\(\bm{u}\cdot\bm{v}=\bm{v}\cdot\bm{u},\)
since multiplication is commutative.
If two vectors are perpendicular (orthogonal)
then \(\bm{u}\cdot\bm{v}=0,\)
and if two vectors are parallel
then \(\bm{u}\cdot\bm{v} = |\bm{u}||\bm{v}|.\)
Intuitively then, the dot product is the product of the magnitudes of two vectors
“correcting for” how much they aren’t pointing in the same direction.
This computation plays an important role in answering the question
“how much of a vector is pointing in the direction of another vector?”
For \(\bm{u} = \langle 3,4\rangle\)
and \(\bm{v} = \langle 5,-2\rangle\)
sketch the projection of \(\bm{u}\) onto \(\bm{v},\)
and sketch the projection of \(\bm{u}\) onto \(\bm{v}^\perp,\)
before defining these things and doing calculations.
For \(\bm{u} = \langle 3,4\rangle\)
and \(\bm{v} = \langle 5,-2\rangle\)
compute \(\operatorname{proj}_{\bm{v}}(\bm{u}).\)
The projection of \(\bm{u}\) onto \(\bm{v}\) (a vector),
denoted \(\operatorname{proj}_{\bm{v}}(\bm{u}),\)
is calculated as
\[
\operatorname{proj}_{\bm{v}}(\bm{u})
= \bigl(|\bm{u}|\cos(\theta)\bigr)\bm{\hat{v} }
= \Bigl(\frac{\bm{u}\cdot\bm{v}}{|\bm{v}|}\Bigr)\bm{\hat{v}}
= \Bigl(\frac{\bm{u}\cdot\bm{v}}{|\bm{v}|^2}\Bigr)\bm{v}
\]
Imagine a light shines directly down, perpendicularly, onto \(\bm{v};\)
\(\operatorname{proj}_{\bm{v}}(\bm{u})\) is the “shadow” that \(\bm{u}\) casts onto \(\bm{v}.\)
The component of \(\bm{u}\) along \(\bm{v}\) (a number)
is the length of the projection of \(\bm{u}\) onto \(\bm{v},\) i.e.
\[
\bigl|\operatorname{proj}_{\bm{v}}(\bm{u})\bigr|
= \frac{\bm{u}\cdot\bm{v}}{|\bm{v}|}
= |\bm{u}|\cos(\theta)
\]
Desmos Demo for the next part.
For a vector \(\bm{v},\) let \(\bm{v}^\perp\) (“vee perp”)
denote the vector perpendicular (orthogonal) to \(\bm{v}\)
and of the same magnitude such that, following the right-hand rule,
the counterclockwise angle from \(\bm{v}\) to \(\bm{v}^\perp\) is 90°.
In terms of its components, if \(\bm{v} = \langle v_1,v_2\rangle\)
then \(\bm{v}^\perp = \langle -v_2,v_1\rangle.\)
We can resolve a vector \(\bm{u}\)
with respect to a coordinate system defined by \(\bm{v}\) and \(\bm{v}^\perp\)
by writing it as a sum of orthogonal vectors,
\[
\bm{u}
= \bigl(|\bm{u}|\cos(\theta)\bigr)\bm{\hat{v}} + \bigl(|\bm{u}|\sin(\theta)\bigr)\bm{\hat{v}}^\perp
= \Bigl(\operatorname{proj}_{\bm{v}}(\bm{u})\Bigr) + \Bigl(\operatorname{proj}_{\bm{v}^\perp}(\bm{u})\Bigr)
= \Bigl(\operatorname{proj}_{\bm{v}}(\bm{u})\Bigr) + \Bigl(\bm{u} - \operatorname{proj}_{\bm{v}}(\bm{u})\Bigr)
\,.
\]
For \(\bm{u} = \langle 3,4\rangle\)
and \(\bm{v} = \langle 5,-2\rangle\)
compute \(\operatorname{proj}_{\bm{v}^\perp}(\bm{u}).\)
Three-Dimensional Rectangular Geometry
Large one-day lecture, then pop quiz.
In three-dimensional space, we add a \(z\)-axis
perpendicular (orthogonal) to the usual \(x\)- and \(y\)-axis
to create a rectangular coordinate system called \(xyz\)-space.
The \(z\)-axis is oriented relative to the \(xy\)-plane
in accordance to the right-hand rule.
Any pair of axes span a coordinate plane,
and these coordinate planes split \(xyz\)-space into eight octants.
Each point can be described by a ordered triple \((x,y,z).\)
Plot the points
\((-1,0,0)\) and \((0,3,0)\) and \((0,0,4)\)
and \((1,2,3)\) and \((-1,4,-5)\)
Given two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2),\)
the distance between them is \( \sqrt{(x_2\!-\!x_1)^2+(y_2\!-\!y_1)^2+(z_2\!-\!z_1)^2}, \)
and their midpoint has coordinates \((\overline{x}, \overline{y}, \overline{z})\)
given by the formulas
\( \overline{x} = \frac{1}{2}(x_1\!+\!x_2) \)
and \( \overline{y} = \frac{1}{2}(y_1\!+\!y_2) \)
and \( \overline{z} = \frac{1}{2}(z_1\!+\!z_2).\)
What’s the distance between the points
\((1,2,3)\) and \((-1,4,-5)?\)
What’s the midpoint between the points
\((1,2,3)\) and \((-1,4,-5)?\)
What’s the distance between the point
\((1,2,3)\) and the generic point \((x,y,z)?\)
What’s an equation that represents all points \((x,y,z)\)
such that the distance between
\((1,2,3)\) and \((x,y,z)\) is seven?
A sphere of radius \(r\) centered at \((h,k,\ell)\)
has equation \( r^2 = (x-h)^2+(y-k)^2+(z-\ell)^2\,. \)
In three-dimensional space,
two points uniquely determine a line
and three non-colinear points uniquely determine a plane.
Generically, a single linear equation determines a plane.
Sketch the planes
\(x=4\) and \(y=-2\) and \(z=7\) and \(x=z.\)
If a sphere intersects a plane in space,
the resulting shape of their intersection, generically a circle,
is called the trace of the sphere in the plane.
What is the radius of the trace of the sphere
\( 7^2 = (x-1)^2+(y-2)^2+(z-3)^2 \)
in the plane \(x=4?\)
Vectors in \(xyz\)-space have three components.
The magnitude of \(\bm{v} = \langle v_1, v_2, v_3\rangle\) is calculated as
\(|\bm{v}| = \sqrt{v_1^2+v_2^2+v_3^2}\,.\)
In addition to the unit coordinate vectors
\({\mathbf{i} = \langle 1,0,0 \rangle}\)
and \({\mathbf{j} = \langle 0,1,0 \rangle,}\)
we now have \({\mathbf{k} = \langle 0,0,1 \rangle}.\)
For vectors \({\bm{u} = \langle u_1,u_2,u_3 \rangle}\)
and \({\bm{v} = \langle v_1,v_2,v_3 \rangle}\)
their dot product is calculated as
\({\bm{u}\cdot\bm{v} = u_1v_1 + u_2v_2 + u_3v_3} = |\bm{u}||\bm{v}|\cos(\theta).\)
Let \(\bm{u} = \langle 1,2,3 \rangle\)
and \(\bm{v} = \langle -1,4,-5 \rangle.\)
Plot them.
What is their sum?
What are their lengths?
What are the unit vectors in their directions?
What is their dot product?
What is the angle between these vectors?
What are the projections of one onto the other?
The direction angles of a vector \(\bm{v}\)
are the three angles between \(\bm{v}\) and each of the three coordinate axes.
The cosines of these angles are referred to as the vector’s direction cosines.
If \(\alpha\) and \(\beta\) and \(\gamma\) are the direction angles of \(\bm{v}\)
with the \(x\)- and \(y\)- and \(z\)-axis respectively, then
\[
\cos(\alpha) = \frac{v_1}{|\bm{v}|}
\quad \cos(\beta) = \frac{v_2}{|\bm{v}|}
\quad \cos(\gamma) = \frac{v_2}{|\bm{v}|}
\qquad \bm{v} = \Bigl\langle |\bm{v}|\cos(\alpha), |\bm{v}|\cos(\beta), |\bm{v}|\cos(\gamma) \Bigr\rangle \,.
\]
Let \(\bm{u} = \langle 1,2,3 \rangle.\)
What are the direction angles of \(\bm{v}?\)
Express \(\bm{v}\) in terms of its direction cosines.
Vectors in Three-Dimensional Space
TK split from previous section
The Cross-Product & Areas
Two-day lecture … but the second day is a pop quiz
Idea: teach all of this without saying “determinant” at all.
Instead just focus on the computation as a computation, no matrices,
and teach it my repeating the vectors in a row
and taking the components to be the forward-product minus the backwards-product.
Give two vectors in space,
there must be a unique direction
that is normal/orthogonal to both of them
… but how do we actually compute it?
A matrix is an two-dimensional array of numbers.
As examples:
A square matrix has the same number of rows as columns.
The determinant is an operation
we can apply to square matrices
that distills the matrix into a number.
I’m going to show you how to compute determinants
of \(2\times 2\) and \(3 \times 3\) matrices,
but note that, for this class, it’s not important in its own right,
but only a means to the end of computing a cross product of vectors.
Compute the determinant of the following \(2 \times 2\) matrices.
Generically the resulting vector will be orthogonal to both \(\bm{u}\) and \(\bm{v}\)
and will point in the direction indicated by the right-hand rule.
It’s length will be proportional to the sine of the angle between \(\bm{u}\) and \(\bm{v}.\)
Specifically, the magnitude of the cross product will be
the product of the vectors’ individual magnitudes
weighted by the sine of the angle between them:
\(|\bm{u} \times \bm{v}| = |\bm{u}||\bm{v}|\sin(\theta).\)
Not coincidentally, this number also equals the area of the parallelogram
(twice the area of the triangle) framed by \(\bm{u}\) and \(\bm{v}\).
As a special case, two vectors \(\bm{u}\) and \(\bm{v}\) are parallel
if and only if \(\bm{u} \times \bm{v} = \bm{0}.\)
What is the area of the triangle in space
with vertices located at the coordinates
\((1,2,3)\) and \((-1,4,-5)\) and \((2,-2,3)?\)
Given a plane in three-dimensional space
a vector \(\bm{n}\) is normal to the plane
if it is orthogonal to every vector parallel to the plane;
i.e. \(\bm{n}\cdot\bm{v}=0\)
for every vector \(\bm{v}\) parallel to the plane.
Three non-colinear points in space uniquely determine a plane.
Given three such points \(P\) and \(Q\) and \(R\)
we can calculate a vector normal to that plane as
\(\overrightarrow{PQ}\times\overrightarrow{PR}.\)
What is a vector normal to the plane
containing the three points
\((-1,4,-5)\) and \((2,3,1)\) and \((1,-4,1)?\)
The scalar triple product
of vectors \(\bm{u}\) and \(\bm{v}\) and \(\bm{w}\)
is calculated as
\[
\bm{u} \cdot \bigl(\bm{v} \times \bm{w}\bigr)
= \operatorname{det}\!\begin{pmatrix}
u_1 & u_2 & u_3
\\ v_1 & v_2 & v_3
\\ w_1 & w_2 & w_3
\end{pmatrix}.
\]
and gives the volume of the parallelepiped,
the three-dimensional analog of a parallelogram,
(and twice the volume of the tetrahedron)
framed by those three vectors.
Specifically, the vectors \(\bm{u}\) and \(\bm{v}\) and \(\bm{w}\)
all lie in the same plane (are coplanar)
if and only if \(\bm{u}\cdot\bigl(\bm{v}\times\bm{w}\bigr)=0.\)
What is the volume of the tetrahedron
with vertices located at the points with coordinates
\((-1,4,-5)\) and \((2,3,1)\) and \((1,-4,1)\) and \((1,3,-1)?\)
Lines and Planes in Three-Dimensional Space
What is the “algebra” underlying a line or plane in space?
Just like in previous math classes we spent a lot of time
studying lines and linear equation in the \(xy\)-plane,
we’ll do the same in this class,
but now there are two different linear geometric features:
1-dimensional lines and 2-dimensional planes.
A plane doesn’t have a single slope anymore;
it depends on what direction you’re headed.
Now though we characterize a plane by its normal vector.
This is not new though:
Consider the standard form
of the line \(y = \frac{3}{4}x+\frac{15}{4},\)
which is \(4y-3x=15.\)
A plane in space passing through a point \(P = (x_0, y_0, z_0)\)
with normal vector \(\bm{n} = \langle A, B, C \rangle\)
will be the set of all points \((x,y,z)\)
such that \(\bm{n} \cdot \bigl(\langle x,y,z \rangle-P\bigr) = 0.\)
Expanding this we get the equation
\({A(x-x_0) + B(y-y_0) + C(z-z_0) = 0,}\)
which can be written as \({Ax + By + Cz = D.}\)
What’s an equation of the plane
with normal vector \(\langle 3,2,-7\rangle\)
that contains the point \((4,-5,1)?\)
Write the equation as \({Ax + By + Cz = D.}\)
What are other points on this plane?
What’s an equation of the plane
containing the three points
\((-1,4,-5)\) and \((2,3,1)\) and \((1,-4,1)?\)
A line in three dimensional space must be described
either as the intersection of two planes
or as a “vector equation” of a parameter \(t.\)
The line passing through a point \(P = (x_0,y_0,z_0)\)
in the direction of the vector \(\bm{v} = \langle v_1,v_2,v_3 \rangle\)
can be described by the parametric equation
\[P + \bm{v}t = \bigl\langle x_0+v_1t, y_0+v_2t, z_0+v_3t \bigr\rangle.\]
What’s an equation of the line
that passes through the points \((0,-3,7)\) and \((1,-1,4)?\)
Does it matter which starting point we use?
It doesn't matter, but just changes the point
a value of \(t\) corresponds to.
Two non-parallel planes in space will intersect along a line.
If the planes have normal vectors \(\bm{n}_1\) and \(\bm{n}_2\),
then the direction vector of the line will be \(\bm{n}_1 \times \bm{n}_2,\)
and any solution to the system of equations consisting of the planes
will result in a point on the line.
What is a parameterization of the line along which
the planes \(2x-4y+z=1\) and \(5x-3y-z=2\) intersect?
A plane in space and a line not parallel to that plane must intersect it at a point.
For the plane \(\bm{n} \cdot \bigl(\langle x,y,z \rangle-P\bigr) = 0\)
and line \({\bigl\langle x_0 + v_1t, y_0 + v_2t, z_0 + v_3t \bigr\rangle}\)
letting \({x = x_0 + v_1t}\) and \({y = y_0 + v_2t}\) and \({z = z+0 + v_3t}\)
in the equation of the plane and solving will result in the value of \(t\)
at which the line hits the plane, which,
when plugged back into the equation of the line itself
will result in the coordinates of the intersection point.
What is the point at which the line
\(\bigl\langle 2,3,-1\bigr\rangle t + (4,1,0)\)
intersects the plane \(2x-4y+z=1?\)
Distances Between Points, Lines, and Planes
Given two geometric objects — points, lines, planes, etc —
how do we compute the (shortest) distance between them?
Compute the shortest distance to a plane from a point not on that plane.
This idea of projecting ends up being the key.
Given a plane in space with normal vector \(\bm{n}\) and containing a point \(Q,\)
and given a point \(P\) not on that plane,
if \(\overrightarrow{QP} = \bm{v},\)
the shortest distance from the point \(P\) to the plane will be
\(\bigl|\operatorname{proj}_{\bm{n}}(\bm{v})\bigr|.\)
What’s the shortest distance from the point \(\bigl(3,-3,10\bigr)\)
to the plane \({2x-3y+6z=-23?}\)
What are the coordinates of the point on that plane closest to that point?
Note \((-1,1,-3)\) lies on the plane.
The distance is \(14\) and the point is \((-1,3,-2).\)
When devising “nice” examples,
start with a Pythagorean quadruple \(\bigr(v_1, v_2, v_3, d\bigl)\)
and a non-nice point \(R\) in space.
For any permutation \(\sigma\) and constant \(c\)
travel away from \(R\) along a vector
\({\bm{n} = c\bigr\langle \pm v_{\sigma 1}, \pm v_{\sigma 2}, \pm v_{\sigma 3}\bigl\rangle}\)
with terminal point \(P\)
and build the equation of the plane passing through \(R\)
with normal vector \(\bm{n}.\)
If the length needs to be a specific number
find a Pythagorean quadruple with diagonal
a divisor of that number, and scale it up.
Compute the shortest distance to a line from a point not on that line.
Call the point \(P,\) and let \(R\) be the point on the line closest to \(P.\)
Let \(\bm{v}\) be the direction vector of the line
and let \(Q\) be the coordinates of any point on the line,
both of which can be inferred from an parameterization of the line.
Let \(\theta\) be the angle between \(\bm{u}\) and \(\bm{v}.\)
which can be computed as
\( \theta = \operatorname{arccos}\bigl(\frac{\bm{u}\cdot\bm{v}}{|\bm{u}||\bm{v}|}\bigr)\,. \)
Considering the right triangle \(\triangle PQR,\)
the distance from \(P\) to the plane —
the distance between \(P\) to \(R\) —
will be \(|\bm{u}\sin\bigl(\theta\bigr)|.\)
What’s the shortest distance from the point \(\bigl(1,-1,-1\bigr)\)
to the line \(\bigl\langle 3,1,4 \bigr\rangle t + \bigl(6,6,0\bigr)?\)
What are the coordinates of the point on that line
closest to the point?
It's \(7,\) and the closest point is \((3,5,-4).\)
Devising nice examples can be done similarly to the case
of “distance between points and planes” problems,
but instead any line in the plane
through \(R\) needs to be determined.
Compute the shortest distance between two skew (non-intersecting) lines.
Let \(P + \bm{w}t\) and \(Q + \bm{v}t\)
be parameterizations of the two lines.
If the lines are parallel,
i.e. \(\bm{w}\) and \(\bm{v}\) are parallel,
then the distance between the lines will be
the distance from the point \(P\) to the line \(Q + \bm{v}t.\)
Otherwise for each line,
there is a unique plane containing that line
that doesn’t intersect the other line.
I.e. there is a pair of parallel planes containing those lines.
Each plane will have normal vector \(\bm{n} = \bm{w}\times\bm{v},\)
and the distance between the lines
will be the distance from \(Q\) to the plane containing \(P\)
(or vice-versa).
What’s the shortest distance between the lines
\({\bigl\langle 0,-2,1 \bigr\rangle t + \bigl(3,0,2\bigr)}\)
and \({\bigl\langle 0,-2,1 \bigr\rangle t + \bigl(0,3,11\bigr)}?\)
It’s \(TK\)
What’s the shortest distance between the lines
\({\bigl\langle 0,-2,1 \bigr\rangle t + \bigl(3,0,2\bigr)}\)
and \({\bigl\langle 4,3,-2 \bigr\rangle t + \bigl(0,3,11\bigr)}?\)
It’s \(9.\)
For each of these types of problems,
let \(R\) denote the point on the line or plane closest to \(P.\)
All of these problems could also be solved
by finding the coordinates of the \(R\)
as the intersection of some line and plane
and computing the length of \(\overline{PR}.\)