Calculus: The Language of Change

Exercise Set 15.8
Geometry and Algebra Lexicon

solutions by Randall Childs and David Theobald

Where do these fancy graphs come from?

The graphs in this notebook were drawn with the aid of the special commands defined in several of the notebooks located under Chapter 15 in the MathmCTLC directory.  The Mathematica gymnastics used to define these special commands are too complicated to explain.  In fact, even the solutions author does not understand them.  Suffice it to say that they take a large set of simpler Mathematica graphics commands and combines them so that the user may draw useful pictures of vectors without crunching a lot of computer programming code.

Exercise 1

Custom Uses of the Length Formula

Compute the length of the vectors

Y = (({{4}, {({{-4}, {2}})}}))	and	Z = (({{2}, {({{-2}, {1}})}}))

|Y| = (Sqrt[(4^2)  +  (((-4))^2)  +  (2^2)]) = (Sqrt[36]) = 6

|Z| = (Sqrt[(2^2)  +  (((-2))^2)  +  (1^2)]) = (Sqrt[9]) = 3

Find a scalar c so that Z = c Y

Z = c Y if and only if c = 1/2.

\(TextForm
    \`\(\(Give\ your\ own\ personal\ formula\ for\ the\ unit\ length\ vector\ 
      that\ points\ in\ the\ same\ direction\ as\ a\ given\ general\ vector, 
    \)
	\)\)

X = (({{(Subscript[x, 1])}, {({{(Subscript[x, 2])}, {(Subscript[x, 3])}})}}))

\(TextForm
    \`It\ is\ not\ essential\ that\ you\ add\ this\ formula\ to\ your\ 
      personal\ lexicon, \ 
    but\ you\ may\ if\ you\ wish . \ \ It\ is\ essential\ that\ you\ can\ find
      \ the\ unit\ vector . \ \ For\ example, \ 
    test\ your\ formula\ \(on : \)\)

(({{(2/3)}, {({{((-1)/3)}, {(2/3)}})}}))  the unit vector in the direction of   (({{2}, {({{-1}, {2}})}})).

Given any vector X, the formula for the unit length vector (Subscript[u, X]) pointing in the same direction as X is :

(Subscript[u, X]) = (X/(|X|)) =\(TraditionalForm
      \`\((x\_1, x\_2, x\_3)\)\/\@\(x\_1\%2 + \ x\_2\%2\  + \ x\_3\%2\)\)

For example, if X = (2,-1,2) then

(Subscript[u, X]) =\(TraditionalForm
      \`\((2, \(-1\), 2)\)\/\@\(2\^2 + \ \((\(-1\))\)\^2\  + \ 2\^2\)\)= (((2, -1, 2))/(Sqrt[9])) = 1/3 (2,-1,2) = (2/3,-1/3,2/3)

which is the correct unit vector.  See the text.

Exercise 2

Sketch U+V-W given the figure.

This first graph simply shows the three given vectors U,V, and W.

The second graph shows the vector addition U+V.  The red arrow labeled 'V' is the displacement vector parallel to V with its tail at the tip of U.

The third graph shows the vector subtraction of W from U+V.  The unlabeled orange arrow is the displacement vector from W to U+V.  The position vector parallel to this displacement vector is the other orange arrow, and it is properly labeled U+V-W.

The fourth graph shows the path "traced" by the vector operation U+V-W.  From the definitions of vector addition and subtraction, one can clearly see that

U + V - W = U + V + (-W)

Therefore U+V-W can be constructed geometrically by drawing the red and blue arrows show.  The red arrow is the "copy" of V whose tail is at the tip of U, and the blue arrow is the "copy" of -W, or the vector W turned around 180 degrees, whose tail is at the tip of U+V.

Exercise 3

Sketch the vectors P=\(TraditionalForm
      \`\( < U∘V > \)\/\( < U∘U > \)\)U and V-P on the figure below.

By definition, P is the projection vector of V onto U.  The real V-P is the position vector parallel to the displacement vector (the "altitude" of V over U) from P to V.

Exercise 4

Circle

Describe geometrically the set of points that consists of the tips of all 2-D position vectors that have a length of five.  Algebraically, this is the set of vectors X((Subscript[x, 1]), (Subscript[x, 2])) satisfying an equation in (Subscript[x, 1])and (Subscript[x, 2]).  What is the equation?

The set of points that are the tips of all 2-D position vectors that have a length of five is, geometrically, the circle of radius five whose center is at the origin.  Algebraically, this is the set of vectors which satisfy the following equation:

(Subscript[x, 1]^2) + (Subscript[x, 2]^2) = (5^2)

Exercise 5

Sphere

Repeat Exercise 15.8.4 using 3-dimensional position vectors and give the equation for the three coordinates.

The set of points that are the tips of all 3-D position vectors that have a length of five is, geometrically, the sphere of radius five whose center is at the origin.  Algebraically, this is the set of vectors which satisfy the following equation:

(Subscript[x, 1]^2) + (Subscript[x, 2]^2) + (Subscript[x, 3]^2) = (5^2)

Exercise 6

Dots in 2-D, Circle in 3-D

What is the intersection of a plane and a sphere?  What is the intersection of a line and a circle?

(a)	Find and draw one the same axes all 2-D position vectors of the form ((Subscript[x, 1]),7) that have length 25.  Etc.

The first condition is the equation (Subscript[x, 2])= 7.  Geometrically, this means that the tips of the vectors must all lie on the line parallel to the (Subscript[x, 1]) axis and displaced 7 units in the positive (Subscript[x, 2]) direction (the horizontal line 7 units above the origin).  Algebraically the condition that the unknown vector have length 25 can be expressed:

(Subscript[x, 1]^2) + (Subscript[x, 2]^2) = (25^2)

This equation along with (Subscript[x, 2])= 7 form the combined algebraic condition.  If you plug and chug, you should get the result that solution vectors must have (Subscript[x, 1]) coordinate of ±24.  Solutions are (24,7) and (-24,7).

(b)	Find and draw one the same axes all 3-D position vectors of the form ((Subscript[x, 1]),7,(Subscript[x, 3])) that have length 25.  Etc.

The first condition is the equation (Subscript[x, 2])= 7.  Geometrically, this means that the tips of the vectors must all lie on the plane parallel to the (Subscript[x, 1])-(Subscript[x, 3]) plane and displaced 7 units in the positive (Subscript[x, 2]) direction (the vertical plane 7 units to the right of the origin).  Algebraically the condition that the unknown vector have length 25 can be expressed:

(Subscript[x, 1]^2) + (Subscript[x, 2]^2) + (Subscript[x, 3]^2) = (25^2)

The geometric meaning of this equation is that all solution vectors must lie on the surface of the sphere whose radius is 25 centered at the origin.  This equation along with (Subscript[x, 2])= 7 form the combined algebraic condition.  If you plug and chug, you should get the result that solution vectors must satisfy the following system of equations.

(Subscript[x, 2]) = 7
(Subscript[x, 1]^2) + (Subscript[x, 3]^2) = (24^2)

The solution set is the circle of radius 24 centered at the point (0,7,0) and located entirely in the vertical plane 7 units to the right of the origin.  This is the intersection of the plane and sphere described above.  Enter the cell below to see Mathematica draw a picture of the solution set.  You may wish to make the graph larger.

The gray line segments are not part of the solution set.  One of these traces the positive (Subscript[x, 2]) axis, which is also the "axis" of the circle.  The other two gray line segments are perpendicular diameters of the circle drawn to suggest the three-dimensional location of the circle.

\(circle\  = \ 
      ParametricPlot3D[{24  Cos[t], 7, 24  Sin[t]}, {t, 0, 2  π}]; 

    axis\  = \ Graphics3D[{GrayLevel[0.4], Line[{{0, 0, 0}, {0, 25, 0}}]}]; 

    diam1\  = \ 
      Graphics3D[{GrayLevel[0.4], Line[{{0, 7, \(-24\)}, {0, 7, 24}}]}]; 

    diam2\  = \ 
      Graphics3D[{GrayLevel[0.4], Line[{{\(-24\), 7, 0}, {24, 7, 0}}]}]; 

    Show[{circle, axis, diam1, diam2}, 
      PlotRange\  -> \ {{\(-25\), 25}, {\(-25\), 25}, {\(-25\), 25}}, 
      ViewPoint\  -> \ {2, 1, 0.5}]\)

Why does this cell work?

The first four lines take sets of three-dimensional graphical information and assign a name to each set.  The last line tells Mathematica to actually draw or "Show" that graphical information with a set of guidelines telling the computer how to draw the graph.

When one defines a graphical object using the ParametricPlot3D[] command, Mathematica automatically draws a picture of the object using default guidelines.  Thus the first line results in the first graph.  It is a graph of the solution set, but it unfortunately offers little information useful for our purposes.  The mathematics that explain why this parametric plot command draws a circle are part of the material to be covered in Chapter 16.  Suffice it to say that the command tells Mathematica to consider the set of all points in three-space whose second coordinate is 7 and whose first and third coordinates are defined by lying on a circle of radius 24.

The lack of illumination provided by the first graph is why the Show[] command is used.  The result is that Mathematica draws a box showing all points in three-dimensional space for which -25≤x1≤+25, -25≤x2≤+25, and -25≤x3≤+25.  Then it draws the solution set circle and the gray line segments as described.

(c)	Find and draw one the same axes all 3-D position vectors of the form ((Subscript[x, 1]),(Subscript[x, 2]),7).  Etc.

Geometrically, this means that the tips of the vectors must all lie on the plane parallel to the (Subscript[x, 1])-(Subscript[x, 2]) plane and displaced 7 units in the positive (Subscript[x, 3]) direction (the horizontal plane 7 units above the origin).  The equation

(Subscript[x, 1]^2) + (Subscript[x, 2]^2) = 625

would mean that the tips of the vectors would lie on a circle of radius 25 centered at some point on the (Subscript[x, 3]) axis if (Subscript[x, 3]) were fixed.  Since (Subscript[x, 3]) is not restricted, then the vectors may lie on the cylinder of radius 25 and infinite length and centered on the (Subscript[x, 3]) axis.  The combined algebraic condition is:

(Subscript[x, 3]) = 7
(Subscript[x, 1]^2) + (Subscript[x, 2]^2) = (25^2)

The solution set is the circle of radius 25 centered at the point (0,0,7) and located entirely in the horizontal plane 7 units above the origin.  This is the intersection of the plane and cylinder described above.  Enter the cell below to see Mathematica draw a picture of the solution set.  You may wish to make the graph larger.

The gray line segments are not part of the solution set.  One of these traces the positive (Subscript[x, 3]) axis, which is also the "axis" of the circle.  The other two gray line segments are perpendicular diameters of the circle drawn to suggest the three-dimensional location of the circle.

\(circle\  = \ 
      ParametricPlot3D[{25  Cos[t], 25  Sin[t], 7}, {t, 0, 2  π}]; 

    axis\  = \ Graphics3D[{GrayLevel[0.4], Line[{{0, 0, 0}, {0, 0, 25}}]}]; 

    diam1\  = \ 
      Graphics3D[{GrayLevel[0.4], Line[{{0, \(-25\), 7}, {0, 25, 7}}]}]; 

    diam2\  = \ 
      Graphics3D[{GrayLevel[0.4], Line[{{\(-25\), 0, 7}, {25, 0, 7}}]}]; 

    Show[{circle, axis, diam1, diam2}, 
      PlotRange\  -> \ {{\(-25\), 25}, {\(-25\), 25}, {\(-25\), 25}}, 
      ViewPoint\  -> \ {2, 1, 0.5}]\)

Why does this cell work?

The first four lines take sets of three-dimensional graphical information and assign a name to each set.  The last line tells Mathematica to actually draw or "Show" that graphical information with a set of guidelines telling the computer how to draw the graph.

When one defines a graphical object using the ParametricPlot3D[] command, Mathematica automatically draws a picture of the object using default guidelines.  Thus the first line results in the first graph.  It is a graph of the solution set, but it unfortunately offers little information useful for our purposes.  The mathematics that explain why this parametric plot command draws a circle are part of the material to be covered in Chapter 16.  Suffice it to say that the command tells Mathematica to consider the set of all points in three-space whose third coordinate is 7 and whose first and second coordinates are defined by lying on a circle of radius 25.

The lack of illumination provided by the first graph is why the Show[] command is used.  The result is that Mathematica draws a box showing all points in three-dimensional space for which -25≤x1≤+25, -25≤x2≤+25, and -25≤x3≤+25.  Then it draws the solution set circle and the gray line segments as described.

Exercise 7

Sliding the Sphere

(a)	Sketch the position vector P = (1,-2,3).

(b)	Sketch and geometrically describe the set of all position vectors X that point to locations in space ....

Sketch and geometrically describe the set of all position vectors X that point to locations in space that are a distance 3 from the tip of P.

This is the set of all vectors whose tips lie on the surface of the sphere of radius three whose center is P.

In this graph, the blue vector is P, and the red vectors are examples of vectors that satisfy the given condition.  These six vectors lie on the octant of the sphere that is in "front" of, to the "right" of, and "below" P.

(c)	Sketch several vectors (Subscript[V, 1]),(Subscript[V, 2]),(Subscript[V, 3]), ... of length 3 with their tails at the tip of P.

\(P = {1, \(-2\), 3}; 
T1 = {3, 0, 0}; 
T2 = {0, 3, 0}; 

    T3 = {0, 0, 3}; 
U1 = {1, 2, \(-2\)}; 
U2 = {2, 1, \(-2\)}; 

    U3 = {2, 2, \(-1\)}; 
V1 = {1, 2, 2}; 
V2 = {2, 1, 2}; 

    View3D[{Arrow[P, Blue], Arrow[T1, Tail -> P, Red], 
        Arrow[T2, Tail -> P, Red], Arrow[\(-T3\), Tail -> P, Red], 
        Arrow[U1, Tail -> P, Red], Arrow[U2, Tail -> P, Red], 
        Arrow[U3, Tail -> P, Red], Arrow[\(-T1\), Tail -> P, Orange], 
        Arrow[\(-T2\), Tail -> P, Orange], Arrow[T3, Tail -> P, Orange], 
        Arrow[\(-U1\), Tail -> P, Orange], Arrow[\(-U2\), Tail -> P, Orange], 
        Arrow[\(-U3\), Tail -> P, Orange], Arrow[\(-V1\), Tail -> P, Orange], 
        Arrow[\(-V2\), Tail -> P, Orange], Arrow[V1, Tail -> P, Orange], 
        Arrow[V2, Tail -> P, Orange]}]; \)

The blue vector is P, and the red and orange vectors are various of the V's.  The six red ones correspond to the six X's drawn in part (b).

(d)	If you draw V with its tail at P, and the position vector X points to the translated V, then X=?

X = P + V
V = X - P

(e)	Write the vector equation |X-P|=3 in coordinates and show that is is equivalent to ....

|X-P| = 3

|(({{x  -  1}, {({{y  -  (-2)}, {z  -  3}})}}))| = 3

(Sqrt[(((x - 1))^2) + (((y + 2))^2) + (((z - 3))^2)]) = 3

(((x - 1))^2) + (((y + 2))^2) + (((z - 3))^2) = 9

(f)	Express the equation |X-P|=3 in English.

The set of unknown vectors X such that the length of the difference vector from P to X is 3.

(g)	Sketch the set of all points X(x,y,z) that satisfy ....

We can use Mathematica to see that the equations in parts (e) and (g) are equivalent.  Look at the expression on the left hand side of the equation in part (e).  We can use the command Expand[] to "multiply out" the squared binomial terms.

(x - 1)^2  +  (y + 2)^2  +  (z - 3)^2

Expand[%]

In Mathematica the "%" symbol represents the most recent output.

Expansion doesn't change the value of this expression.  It should still be equal to 9.  We can simplify the constant term of this new equation by subtracting 9 from both sides.  The result

(x^2) - 2x + (y^2) + 4y + (z^2) - 6z + 5 = 0

is precisely the equation given in part (g).  Thus the set of points that satisfy this equation is the same set of points that satisfy the vector equation |X-P| = 3.  So we're still talking about the sphere of radius 3 whose center is P.

Exercise 8

Find the angle between two diagonals on adjacent faces of a cube.

Enter the cell below to see Mathematica draw a picture of a cube with edges parallel to the (Subscript[x, 1]), (Subscript[x, 2]), and (Subscript[x, 3]) axes and vertices at the origin and (1,1,1).  You may wish to make the graph larger.  Mathematica doesn't draw the (Subscript[x, 1]), (Subscript[x, 2]), and (Subscript[x, 3]) axes in its 3-D plots, but the graph has been oriented so that the positive and negative directions of (Subscript[x, 1]), (Subscript[x, 2]), and (Subscript[x, 3]) in these plots correspond to the same directions in the 3-D diagrams in the text.

The gray line segments are the edges of the box.  Three of the edges coincide with the positive (Subscript[x, 1]),(Subscript[x, 2]), and (Subscript[x, 3]) axes.  These edges are "behind" the other nine edges and would be invisible if the box were constructed with some opaque material.  These three edges are a lighter gray than the others.  The two black line segments originating at the intersection of the light gray edges are the diagonals we wish to investigate.

The black line segments can be represented by the vectors (1,0,1) and (0,1,1).  From the geometry/algebra lexicon we know the cosine of the angle between two vectors:

Cos[θ] = ((⟨X•Y⟩)/(|X| |Y|))

It is easy to compute that the length of each vector is (Sqrt[2]).  It is also easy to compute that the dot product of the two vectors is 1.  Thus the cosine of the angle between the diagonals is 1/2, and the angle itself is 60 degrees or π/3 radians.

\(AB\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{1, 0, 0}, \ {1, 1, 0}}]\ }\ ]; 
    
BC\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{1, 1, 0}, \ {0, 1, 0}}]\ }\ ]; 
    
EF\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{1, 0, 1}, \ {1, 1, 1}}]\ }\ ]; 
    
FG\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{1, 1, 1}, \ {0, 1, 1}}]\ }\ ]; 
    
EH\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{1, 0, 1}, \ {0, 0, 1}}]\ }\ ]; 
    
GH\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{0, 1, 1}, \ {0, 0, 1}}]\ }\ ]; 
    
AE\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{1, 0, 0}, \ {1, 0, 1}}]\ }\ ]; 
    
BF\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{1, 1, 0}, \ {1, 1, 1}}]\ }\ ]; 
    
CG\  = \ 
      Graphics3D[\ {\ GrayLevel[0.4], \ Line[{{0, 1, 0}, \ {0, 1, 1}}]\ }\ ]; 
    
DA\  = \ 
      Graphics3D[\ {\ GrayLevel[0.8], \ Line[{{0, 0, 0}, \ {1, 0, 0}}]\ }\ ]; 
    
DC\  = \ 
      Graphics3D[\ {\ GrayLevel[0.8], \ Line[{{0, 0, 0}, \ {0, 1, 0}}]\ }\ ]; 
    
DH\  = \ 
      Graphics3D[\ {\ GrayLevel[0.8], \ Line[{{0, 0, 0}, \ {0, 0, 1}}]\ }\ ]; 
    
de\  = \ Graphics3D[Line[{{0, 0, 0}, {1, 0, 1}}]]; 

    dg = \ Graphics3D[\ Line[{{0, 0, 0}, {0, 1, 1}}]]; 

    Show[{AB, BC, EF, FG, EH, GH, AE, BF, CG, DA, DC, DH, de, dg}, \ 
      Axes\  -> \ True, \ 
      PlotRange\  -> \ {{\(-1\), 1}, {\(-1\), 1}, {\(-1\), 1}}, \ 
      ViewPoint\  -> \ {1.1, 1, 0.9}]; \)

Why does this cell work?

The first fourteen lines take sets of three-dimensional graphical information and assign a name to each set.  The last line tells Mathematica to actually draw or "Show" that graphical information with a set of guidelines telling the computer how to draw the graph.

The result is that Mathematica draws a box showing all points in three-dimensional space for which -1≤x1≤+1, -1≤x2≤+1, and -1≤x3≤+1.  Then it draws gray line segments which are the edges of the smaller box as described above.  The names given to these line segments follow the traditional geometric custom of naming important points in a geometric figure by capital letters and assigning to any line segment the pair of letters corresponding to its endpoints.  Then it draws in black the diagonals which start at the origin and traverse the "back left" and "back right" faces.  These are diagonals of adjacent faces, and they meet at one point, so it makes sense to talk about the angle between these two line segments.