- This exercise is recommended for all readers.
- This exercise is recommended for all readers.
- This exercise is recommended for all readers.
- 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.
![{\displaystyle {\begin{array}{rl}({\vec {u}}+{\vec {v}})\cdot {\vec {w}}&=[{\begin{pmatrix}u_{1}\\\vdots \\u_{n}\end{pmatrix}}+{\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}]\cdot {\begin{pmatrix}w_{1}\\\vdots \\w_{n}\end{pmatrix}}\\&={\begin{pmatrix}u_{1}+v_{1}\\\vdots \\u_{n}+v_{n}\end{pmatrix}}\cdot {\begin{pmatrix}w_{1}\\\vdots \\w_{n}\end{pmatrix}}\\&=(u_{1}+v_{1})w_{1}+\cdots +(u_{n}+v_{n})w_{n}\\&=(u_{1}w_{1}+\cdots +u_{n}w_{n})+(v_{1}w_{1}+\cdots +v_{n}w_{n})\\&={\vec {u}}\cdot {\vec {w}}+{\vec {v}}\cdot {\vec {w}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d3a18a0541df6949cc52527b74a494a8add2ae6)
- 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-Schwartz 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.
- 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.
- This exercise is recommended for all readers.
- Problem 16
Prove that
- Answer
Write

and then this computation works.

- 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-Schwartz 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.
- 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 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 31
Verify the Cauchy-Schwartz 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-Schwartz 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
![{\displaystyle {\begin{aligned}&={\bigl [}(\sum _{1\leq j\leq n}{a_{j}}^{2})+{a_{n+1}}^{2}{\bigr ]}{\bigl [}(\sum _{1\leq j\leq n}{b_{j}}^{2})+{b_{n+1}}^{2}{\bigr ]}\\&\quad -{\bigl [}\sum _{1\leq k<j\leq n}{\bigl (}a_{k}b_{j}-a_{j}b_{k}{\bigr )}^{2}+\sum _{1\leq k\leq n}{\bigl (}a_{k}b_{n+1}-a_{n+1}b_{k}{\bigr )}^{2}{\bigr ]}\\&={\bigl (}\sum _{1\leq j\leq n}{a_{j}}^{2}{\bigr )}{\bigl (}\sum _{1\leq j\leq n}{b_{j}}^{2}{\bigr )}+\sum _{1\leq j\leq n}{b_{j}}^{2}{a_{n+1}}^{2}+\sum _{1\leq j\leq n}{a_{j}}^{2}{b_{n+1}}^{2}+{a_{n+1}}^{2}{b_{n+1}}^{2}\\&\qquad -{\bigl [}\sum _{1\leq k<j\leq n}{\bigl (}a_{k}b_{j}-a_{j}b_{k}{\bigr )}^{2}+\sum _{1\leq k\leq n}{\bigl (}a_{k}b_{n+1}-a_{n+1}b_{k}{\bigr )}^{2}{\bigr ]}\\&={\bigl (}\sum _{1\leq j\leq n}{a_{j}}^{2}{\bigr )}{\bigl (}\sum _{1\leq j\leq n}{b_{j}}^{2}{\bigr )}-\sum _{1\leq k<j\leq n}{\bigl (}a_{k}b_{j}-a_{j}b_{k}{\bigr )}^{2}\\&\quad +\sum _{1\leq j\leq n}{b_{j}}^{2}{a_{n+1}}^{2}+\sum _{1\leq j\leq n}{a_{j}}^{2}{b_{n+1}}^{2}+{a_{n+1}}^{2}{b_{n+1}}^{2}\\&\qquad -\sum _{1\leq k\leq n}{\bigl (}a_{k}b_{n+1}-a_{n+1}b_{k}{\bigr )}^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5af02a164172eb08a2316019847eb6d4244db9e)
and apply the inductive hypothesis
![{\displaystyle {\begin{array}{rl}&={\bigl (}\sum _{1\leq j\leq n}a_{j}b_{j}{\bigr )}^{2}+\sum _{1\leq j\leq n}{b_{j}}^{2}{a_{n+1}}^{2}+\sum _{1\leq j\leq n}{a_{j}}^{2}{b_{n+1}}^{2}+{a_{n+1}}^{2}{b_{n+1}}^{2}\\&\qquad -{\bigl [}\sum _{1\leq k\leq n}{a_{k}}^{2}{b_{n+1}}^{2}-2\sum _{1\leq k\leq n}a_{k}b_{n+1}a_{n+1}b_{k}+\sum _{1\leq k\leq n}{a_{n+1}}^{2}{b_{k}}^{2}{\bigr ]}\\&={\bigl (}\sum _{1\leq j\leq n}a_{j}b_{j}{\bigr )}^{2}+2{\bigl (}\sum _{1\leq k\leq n}a_{k}b_{n+1}a_{n+1}b_{k}{\bigr )}+{a_{n+1}}^{2}{b_{n+1}}^{2}\\&={\bigl [}{\bigl (}\sum _{1\leq j\leq n}a_{j}b_{j}{\bigr )}+a_{n+1}b_{n+1}{\bigr ]}^{2}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e05d3b51a0ab84f0cd5087e6427463c0b87af97)
to derive the left-hand side.
- O'Hanian, Hans (1985), Physics, 1, W. W. Norton
- Ivanoff, V. F. (proposer); Esty, T. C. (solver) (Feb. 1933), "Problem 3529", American Mathematical Mothly 39 (2): 118