Calculus: The Language of Change

Exercise Set 15.6
Angles and Projections

solutions by Randall Childs and David Theobald

See the text for explicit instructions on using a Mathematica notebook to solve Exercises 2 and 3.

Exercise 1

Your Own Condition for Perpendicularity

Give the simplest possible algebraic condition you can use to test whether two vectors are perpendicular.

HINT:  What is Cos[(π/2)]?

For example, are (({{1}, {-3}, {2}})) and (({{2}, {2}, {1}})) perpendicular?

Are  (({{1}, {-3}, {2}})) and (({{1}, {2}, {1}})) ?

What are the angles between them?  Does the denominator matter?

YOUR CONDITION FOR PERPENDICULARITY:  A is perpendicular to B if and only if ????

Two vectors are perpendicular if and only if they form a right angle.  In other words, the angle between them is

(π/2)

We know that Cos[(π/2)] = 0.

So we can restate the perpendicularity condition as:  two vectors are perpendicular if and only if the cosine of the angle between them is zero.

The dot (or inner) product gives us a formula for the cosine of the angle between two vectors.

Cos[θ] = ((⟨X•Y⟩)/(|X||Y|))

where θ is the angle between the vectors X and Y.

So Cos[θ] is zero if and only if ⟨X•Y⟩ is zero.  The value of the denominator, |X||Y|, does not affect this condition.  If the cosine of the included angle is zero, then the product of that with the denominator is still zero.  The only way the denominator could cause trouble is if it were zero.  But that would require either X or Y to be the zero vector, and the idea of an angle between any vector and the zero vector makes no geometric sense.  In order for two vectors to have an included angle, neither may be the zero vector.  Therefore, the denominator is nonzero, and our condition is well-defined.

Condition for Perpendicularity

X is perpendicular to Y if and only if their dot product is zero.

Neither of the two pairs of vectors given by the text is a pair of perpendicular vectors.  The dot product of the first pair is -2, and the dot product of the second pair is -3.  Use the notebook AnglPerp to compute the angles between these pairs of vectors.

Using the condition for perpendicularity, convince yourself that

(({{({{2}, {-3}})}, {2}})) and (({{({{4}, {4}})}, {2}}))

are perpendicular vectors.  Enter the following cell to see a graph of these vectors.  The graph uses a "perfect" vantage point which allows these vectors to actually appear perpendicular.

\(X = {2, \(-3\), 2}; 
Y = {4, 4, 2}; 

    xvect\  = \ Graphics3D[\ Line[{{0, 0, 0}, \ X}]]; 

    vecty\  = \ Graphics3D[\ Line[{{0, 0, 0}, \ Y}]]; 

    Show[{xvect, vecty}, \ Axes\  -> \ True, \ 
      PlotRange\  -> \ {{\(-5\), 5}, {\(-5\), 5}, {\(-5\), 5}}, \ 
      ViewPoint\  -> \ {\(-7\), 2, 10}]; \)

Why does this cell work?

The first two lines give the names X and Y to the given vectors.  The next two take sets of three-dimensional graphical information involving X and Y and assign a name to each set.  The fifth one tells Mathematica to actually draw or "Show" those sets of 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 -5≤x1≤+5, -5≤x2≤+5, and -5≤x3≤+5.  Then it draws line segments representing the vectors X and Y.  The specific numbers chosen in the ViewPoint command allow X and Y to appear perpendicular in our graph (which we must remember is only a two-dimensional model of three-dimensional space).

To see that this is true, paste a copy of the above cell somewhere else in this notebook.  Think of the numbers in the ViewPoint command as a position vector, and replace it with a vector which is not parallel to (-7,2,10).  Exercise 3 from Exercise Set 15.4 gives an easy algebraic condition to determine whether two position vectors are parallel.  If you enter the cell with this new ViewPoint, then X and Y will no longer appear to be perpendicular.

Don't look at this until you've learned about cross products!

Enter the cell below.  It does what it looks like it does: compute the cross (or outer) product of X and Y.  Its output is a familiar vector.  Is this a coincidence?  HINT:  The solutions author asked Mathematica to perform this calculation before choosing a ViewPoint for the graph of X and Y.

X\[Cross]Y