Find the angle between each two, if it is defined.
Answer
Not defined.
This exercise is recommended for all readers.
Problem 3
During maneuvers preceding the Battle of Jutland,
the British battle cruiser Lion moved as follows (in nautical
miles): miles north, miles degrees
east of south, miles at degrees east of
north, and miles at degrees east of north.
Find the distance between starting and ending positions (O'Hanian 1985).
Answer
We express each displacement as a vector (rounded to one
decimal place because that's the accuracy of the problem's statement)
and add to find the total displacement
(ignoring the curvature of the earth).
The distance is .
Problem 4
Find so that these two vectors are perpendicular.
Answer
Solve to get .
Problem 5
Describe the set of vectors in orthogonal to this one.
Answer
The set
can also be described with parameters in this way.
This exercise is recommended for all readers.
Problem 6
Find the angle between the diagonal of the unit square in
and one of the axes.
Find the angle between the diagonal of the unit cube in
and one of the axes.
Find the angle between the diagonal of the unit cube in
and one of the axes.
What is the limit, as goes to ,
of the angle between the diagonal of the unit cube in
and one of the axes?
Answer
We can use the -axis.
Again, use the -axis.
The -axis worked before and it will work again.
Using the formula from the prior item,
.
Problem 7
Is any vector perpendicular to itself?
Answer
Clearly is zero if and only if
each is zero.
So only is perpendicular to itself.
This exercise is recommended for all readers.
Problem 8
Describe the algebraic properties of dot product.
Is it right-distributive over addition:
?
Is is left-distributive (over addition)?
Does it commute?
Associate?
How does it interact with scalar multiplication?
As always, any assertion must be backed by either a proof or an example.
Answer
Assume that have components
.
Dot product is right-distributive.
Dot product is also left distributive:
.
The proof is just like the prior one.
Dot product commutes.
Because
is a scalar, not a vector,
the expression makes no
sense; the dot product of a scalar and a vector is not defined.
This is a vague question so it has many answers.
Some are
(1)
and ,
(2)
(in general; an example is easy to produce), and
(3)
(the connection between
norm and dot product is that the square of the norm is the
dot product of a vector with itself).
Problem 9
Verify the equality condition in Corollary 2.6,
the Cauchy-Schwarz Inequality.
Show that if is a negative scalar multiple of
then and
are less than or equal to zero.
Show that
if and only if one vector is a scalar multiple of the other.
Answer
Verifying that
for and
is easy.
Now, for and ,
if then
,
which is times a nonnegative
real.
The half is similar (actually, taking the
in this paragraph to be the reciprocal of the
above gives that we need only worry about the case).
We first consider the case.
From the Triangle Inequality we know that
if and only if one vector is a nonnegative scalar multiple of the
other.
But that's all we need because the
first part of this exercise shows that,
in a context where the dot product of the two vectors is positive,
the two statements
"one vector is a
scalar multiple of the other" and "one vector is a nonnegative
scalar multiple of the other", are equivalent.
We finish by considering the case.
Because
and
,
we have that
.
Now the prior paragraph applies to give that one of the two vectors
and is a scalar multiple of the other.
But that's equivalent to the assertion that one of the two vectors
and is a scalar multiple of the other,
as desired.
Problem 10
Suppose that
and .
Must ?
Answer
No.
These give an example.
This exercise is recommended for all readers.
Problem 11
Does any vector have
length zero except a zero vector?
(If "yes", produce an example.
If "no", prove it.)
Answer
We prove that a vector has length zero if and only if all its
components are zero.
Let have components .
Recall that the square of any real number is greater than or equal to
zero, with equality only when that real is zero.
Thus
is a sum of numbers
greater than or equal to zero, and so is itself greater than or equal
to zero, with equality if and only if each is zero.
Hence if and only if all the components of
are zero.
This exercise is recommended for all readers.
Problem 12
Find the midpoint of the line segment connecting
with in .
Generalize to .
Answer
We can easily check that
is on the line connecting the two, and is equidistant from both.
The generalization is obvious.
Problem 13
Show that if then
has length one.
What if ?
Answer
Assume that has components .
If then we have this.
If then is not
defined.
Problem 14
Show that if then
is times as long
as .
What if ?
Answer
For the first question, assume that and
, take the root, and factor.
For the second question, the result is times as long, but it
points in the opposite direction in that
.
This exercise is recommended for all readers.
Problem 15
A vector of length one is a
unit vector.
Show that the dot product of two unit vectors has absolute value
less than or equal to one.
Can "less than" happen?
Can "equal to"?
Answer
Assume that both have length .
Apply Cauchy-Schwarz:
.
To see that "less than" can happen, in take
and note that .
For "equal to", note that .
Problem 16
Prove that
Answer
Write
and then this computation works.
Problem 17
Show that if for every
then .
Answer
We will prove this demonstrating that the contrapositive
statement holds: if then there
is a with .
Assume that .
If then it has a nonzero component, say the
-th one .
But the vector that is all zeroes except for
a one in component gives
.
(A slicker proof just considers .)
Problem 18
Is
?
If it is true then it would generalize the Triangle Inequality.
Answer
Yes;
we can prove this by induction.
Assume that the vectors are in some .
Clearly the statement applies to one vector.
The Triangle Inequality is this statement applied to two vectors.
For an inductive step assume the statement is true for or fewer
vectors.
Then this
follows by the Triangle Inequality for two vectors.
Now the inductive hypothesis, applied to the first summand on the right,
gives that as less than or equal to
.
Problem 19
What is the ratio between the sides in the Cauchy-Schwarz inequality?
Answer
By definition
where is the angle between the vectors.
Thus the ratio is .
Problem 20
Why is the zero vector defined to be perpendicular to every vector?
Answer
So that the statement "vectors are orthogonal iff their
dot product is zero" has no exceptions.
Problem 21
Describe the angle between two vectors in .
Answer
The angle between and is found
(for ) with
If or is zero then the angle is radians.
Otherwise, if and are of opposite signs then the angle is
radians, else the angle is zero radians.
Problem 22
Give a simple necessary and sufficient condition to determine
whether the angle between two vectors is acute, right, or obtuse.
Answer
The angle between and is acute
if , is right if
, and is obtuse if
.
That's because, in the formula for the angle, the denominator is never
negative.
This exercise is recommended for all readers.
Problem 23
Generalize to the converse of the Pythagorean
Theorem, that
if and are
perpendicular then
.
Answer
Suppose that .
If and are perpendicular then
(the third equality holds because ).
Problem 24
Show that
if and only if and
are perpendicular.
Give an example in .
Answer
Where , the vectors
and are perpendicular if and
only if
,
which shows that those two are perpendicular if and only if
.
That holds if and only if .
Problem 25
Show that if a vector is perpendicular to each of two others then
it is perpendicular to each vector in the plane they generate.
(Remark.
They could generate a
degenerate plane— a line or a point— but
the statement remains true.)
Answer
Suppose is perpendicular to both
and .
Then, for any we have this.
Problem 26
Prove that, where are nonzero vectors,
the vector
bisects the angle between them.
Illustrate in .
Answer
We will show something more general: if
for
, then
bisects the angle between and
(we ignore the case where and are
the zero vector).
The case is easy.
For the rest, by the definition of angle,
we will be done if we show this.
But distributing inside each expression gives
and , so the
two are equal.
Problem 27
Verify that the definition of angle is dimensionally correct:
(1) if then the cosine of the angle between
and equals the cosine of the angle between
and , and (2) if then the cosine of the angle
between and is the negative of the cosine
of the angle between and .
Answer
We can show the two statements together.
Let , write
and calculate.
This exercise is recommended for all readers.
Problem 28
Show that the inner product operation is linear: for
and ,
.
Answer
Let
and then
as required.
This exercise is recommended for all readers.
Problem 29
The geometric mean
of two positive reals is .
It is analogous to the arithmetic mean.
Use the Cauchy-Schwarz inequality to show that
the geometric mean of any is less
than or equal to the arithmetic mean.
Answer
For , set
so that the Cauchy-Schwarz inequality asserts that (after squaring)
as desired.
? Problem 30
A ship is sailing with speed and direction ; the wind blows
apparently (judging by the vane on the mast) in the direction of a vector
; on changing the direction and speed of the ship from
to the apparent wind is in the direction
of a vector .
This is how the answer was given in the cited source.
The actual velocity of the wind is the sum of the
ship's velocity and the apparent velocity of the wind.
Without loss of generality we may assume and
to be unit vectors, and may write
where and are undetermined scalars.
Take the dot product first by and then by
to obtain
Multiply the second by ,
subtract the result from the first, and find
Substituting in the original displayed equation, we get
Problem 31
Verify the Cauchy-Schwarz inequality by first proving
Lagrange's identity:
and then noting that the final term is positive.
(Recall the meaning
and
of the notation.)
This result
is an improvement over Cauchy-Schwarz because it gives a formula for
the difference between the two sides.
Interpret that difference in .
Answer
We use induction on .
In the base case the identity reduces to
and clearly holds.
For the inductive step assume that
the formula holds for the , ..., cases.
We will show that it then holds in the case.
Start with the right-hand side