Solution to Homework 1
Olena Bormashenko
September 11, 2011
Section 1.1: 4, 5(b)(d), 7(b)(f), 8(a)(b), 22; Section 1.2: 2, 5, 7, 9, 10,
11(b)(c), 12, 15(b), 23
Section 1.1
4. In each of the following cases, find a point that is two-thirds of the distance
from the first (initial) point to the second (terminal) point.
(a) (−4, 7, 2), (10, −10, 11)
Solution: Given two points P and Q, the vector that represents
the movement from P to Q is P − Q. We are looking for a point A
which is two thirds of the distance from P to Q – this is equivalent to
saying that we start at P , and go two-thirds of the way to Q. Going
all the way to Q corresponds to P − Q – this means that going to A
corresponds to going 23 (P − Q).
In this particular case, we see that the vector corresponding to the
motion from the first point to the second point is (10, −10, 11) −
(−4, 7, 2) = [14, −17, 9]. This means that the vector corresponding
to going two-thirds of the way is
2
[14, −17, 9] = [28/3, −34/3, 6]
3
Therefore, to get from P to A we need to move according to the
above vector. Thus, A − P = [28/3, −34/3, 6], and so
A = P + [28/3, −34/3, 6] = (−4, 7, 2) + [28/3, −34/3, 6]
= (16/3, −13/3, 8)
(b) (2, −1, 0, −7), (−11, −1, −9, 2).
Solution: Doing the same kind of calculation as in part (a), we
see that our point is precisely:
(2, −1, 0, −7) +
2
((−11, −1, −9, 2) − (2, −1, 0, −7))
3
1
which simplifies to (−20/3, −1, −6, −1) .
5. In each of the following cases, find a unit vector in the same direction
as the given vector. Is the resulting (normalized)vector longer or shorter
than the original? Why?
(b) [4, 1, 0, −2]
Solution: In order to find a unit vector in the same direction as
a given vector, just divide by the length of the vector. (This makes
intuitive sense: if you had a vector of length 2, then to get a unit
vector (a vector of length 1) in the same direction, you would divide
by 2.) Furhermore, it’s clear that our unit vector is longer than the
original vector if and only if the vector we started with had length
less than 1. Let ~u be our unit vector. Then,
[4, 1, 0, −2]
[4, 1, 0, −2]
=p
k[4, 1, 0, −2]k
42 + 12 + 02 + (−2)2
4
1
2
= √ , √ , 0, − √
21
1
21
√
Since the length of the original vector 21, the resulting unit vector
is shorter than the original vector.
(d) 15 , − 25 , − 15 , 15 , 25
~u =
Solution: In the same way as before,
1 2 1 1 2
5, −5, −5, 5, 5
~u = q
2
2
2
2
2
− 25 + − 15 + 15 + 15 + 25
1 2 1 1 2
1 2 1 1 2
,− ,− , ,
5, −5, −5, 5, 5
q
=
= 5 5√ 5 5 5
11
5
11
25
2
1
1
2
1
√ , −√ , −√ , √ , √
11
11
11
11
11
q
Since the length of the original vector is 11
25 , the normalized vector
=
is longer than the original vector.
7. If ~x = [−2, 4, 5], ~y = [−1, 0, 3], and ~z = [4, −1, 2], find the following
(b) −2~y
Solution: By definition,
−2~y = −2[−1, 0, 3] = [2, 0, −6]
2
(f) 2~x + 3~y − 4~z
Solution: By definition,
2~x + 3~y − 4~z = 2[−2, 4, 5] + 3[−1, 0, 3] − 4[4, −1, 2]
= [−4, 8, 10] + [−3, 0, 9] − [16, −4, 8]
= [−23, 12, 11]
8. Given ~x and ~y as follows, calculate ~x + ~y , ~x − ~y , and ~y − ~x and sketch
~x, ~y , ~x + ~y , ~x − ~y in the same coordinate system.
(a) ~x = [−1, 5], ~y = [2, −4]
Solution:
~x + ~y = [−1, 5] + [2, −4] = [1, 1]
~x − ~y = [−1, 5] − [2, −4] = [−3, 9]
~y − ~x = [2, −4] − [−1, 5] = [3, −9]
(b) ~x = [10, −2], ~y = [−7, −3]
Solution:
~x + ~y = [10, −2] + [−7, −3] = [3, −5]
~x − ~y = [10, −2] − [−7, −3] = [17, 1]
~y − ~x = [−7, −3] − [10, −2] = [−17, −1]
22. (a) Prove that the length of each vector in Rn is nonnegative.
Proof:
Assumptions: ~x is a vector in Rn .
Need to show: k~xk ≥ 0.
Let ~x = [x1 , x2 , . . . , xn ]. Then, k~xk =
squares are nonnegative,
p
x21 + x22 + · · · + x2n . Since
x21 + x22 + · · · + x2n ≥ 0
Since the square root of a nonnegative
number is defined to be nonp
negative, we see that k~xk = x21 + x22 + · · · + x2n ≥ 0, so we’re done.
(b) Prove that the only vector in Rn of length 0 is the zero vector.
3
Proof:
Assumptions: k~xk = 0
Need to show: ~x = ~0
Let ~x = [x1 , x2 , . . . , xn ]. Then, we have that
q
0 = k~xk = x21 + · · · + x2n
⇒ 0 = x21 + · · · + x2n
Since each of the x2i is non-negative, and their sum is 0, the only way
this is possible is that each is 0. Thus, x2i is 0 for each i, and hence
~x = ~0.
Section 1.2
2. Show that the points A1 (9, 19, 16), A2 (11, 12, 13), and A3 (14, 23, 10) are
the vertices of a right triangle. (Hint: Construct vectors between the
points and check for an orthogonal pair.)
Solution: The vectors that represent the sides of the triangle are A1 −
A2 , A2 −A3 , and A1 −A3 . (We had to choose between A1 −A2 and A2 −A1 ,
etc., but this is not important.) Accordingly, they are:
A1 − A2 = (9, 19, 16) − (11, 12, 13) = [−2, 7, 3]
A2 − A3 = (11, 12, 13) − (14, 23, 10) = [−3, −11, 3]
A3 − A1 = (14, 23, 10) − (9, 19, 16) = [5, 4, −6]
Checking, the various pairs, we see that
(A1 − A2 ) · (A2 − A3 ) = [−2, 7, 3] · [−3, −11, 3] = −62
(A2 − A3 ) · (A3 − A1 ) = [−3, −11, 3] · [5, 4, −6] = −77
(A3 − A1 ) · (A1 − A2 ) = [5, 4, −6] · [−2, 7, 3] = 0
Thus, the last equation shows that (A3 −A1 ) is perpendicular to (A1 −A2 ),
and therefore the triangle has a right angle at A1 .
5. Why isn’t it true that if ~x, ~y , ~z ∈ Rn , then ~x · (~y · ~z) = (~x · ~y ) · ~z?
Solution:
Since ~y · ~z is a scalar, the dot product ~x · (~y · ~z) is simply not defined – it’s
impossible to calculate the dot product of a vector and a scalar. Therefore,
neither left-hand side nor right-hand side are defined, so the statement is
not true.
7. Does the Cancellation Law of algebra always hold for the dot product:that
is, assuming that ~z 6= 0, does ~x · ~z = ~y · ~z always imply that ~x = ~y ?
4
Solution: This does not hold. The simplest example one can come up
with is an example where ~z is perpendicular to both ~y and ~x – in that case,
both the dot products are 0. For example, let ~x = [0, 1, 1], ~y = [0, 1, 0] and
let ~z = [1, 0, 0]. In that case,
~x · ~z = 0 = ~y · ~z
but ~x 6= ~y .
9. Prove that if (~x + ~y ) · (~x − ~y ) = 0 then k~xk = k~y k.
Proof:
Assumptions: (~x + ~y ) · (~x − ~y ) = 0
Need to show: k~xk = k~y k
As usual, we use the assumption to show the desired result. Here, expanding things out,
0 = (~x + ~y ) · (~x − ~y ) = ~x · ~x − ~y · ~x + ~x · ~y − ~y · ~y
= ~x · ~x − ~y · ~y = k~xk2 − k~y k2
using the fact that ~x · ~y = ~y · ~x and the identity for length in terms of the
dot product. Rearranging, this yields k~xk2 = k~y k2 . Since the lengths are
both nonnegative, we can take the square roots of both sides to conclude
that k~xk = k~y k, as required.
2
2
2
2
10. Prove that 12 k~x + ~y k + k~x − ~y k = k~xk + k~y k .
Proof:
Assumptions: Noreal assumptions. 2
2
2
2
Need to show: 21 k~x + ~y k + k~x − ~y k = k~xk + k~y k
Here, there are no assumptions, so we just need to check that the lefthand side and the right-hand side are equal in general. Since we have
squares of lengths, we rewrite everything in terms of dot products. We
start from the left-hand side and manipulate it:
1
1
2
2
k~x + ~y k + k~x − ~y k = ((~x + ~y ) · (~x + ~y ) + (~x − ~y ) · (~x − ~y ))
2
2
1
= (~x · ~x + 2~x · ~y + ~y · ~y + ~x · ~x − 2~x · ~y + ~y · ~y )
2
1
= (2~x · ~x + 2~y · ~y )
2
= k~xk2 + k~y k2
as required.
5
11. (b) Prove that if ~x, ~y , ~z are mutually orthogonal vectors in Rn , then
2
2
2
2
k~x + ~y + ~zk = k~xk + k~y k + k~zk
Proof:
Assumptions: ~x, ~y and ~z are mutually orhogonal: that is, ~x · ~y =
0, ~x · ~z = 0, and ~y · ~z = 0
2
2
2
2
Need to show: k~x + ~y + ~zk = k~xk + k~y k + k~zk
Again, rewriting everything in terms of dot products:
2
k~x + ~y + ~zk = (~x + ~y + ~z) · (~x + ~y + ~z)
= ~x · ~x + ~y · ~y + ~z · ~z + 2~x · ~y + 2~y · ~z + 2~x · ~z
= ~x · ~x + ~y · ~y + ~z · ~z
= k~xk2 + k~y k2 + k~zk2
using the assumption, and the identity for the length of a vector.
2
2
(c) Prove that ~x · ~y = 41 k~x + ~y k − k~x − ~y k .
Proof:
Assumptions: None.
Need to show: ~x · ~y =
1
4
2
2
k~x + ~y k − k~x − ~y k
.
Using dot products once again, and starting from the right-hand
side:
1
1
2
2
k~x + ~y k − k~x − ~y k = ((~x + ~y ) · (~x + ~y ) − (~x − ~y ) · (~x − ~y ))
4
4
1
= (~x · ~x + 2~x · ~y + ~y · ~y − (~x · ~x − 2~x · ~y + ~y · ~y ))
4
1
= (4~x · ~y ) = ~x · ~y
4
as required.
12. Given ~x, ~y , ~z in Rn , with ~x orthogonal to both ~y and ~z, prove that ~x is
orthogonal to c1 ~y + c2 ~z where c1 , c2 ∈ R.
Proof:
Assumptions: ~x orthogonal to both ~y and ~z: that is, ~x ·~y = 0 and ~x ·~z = 0.
Need to show: ~x · (c1 ~y + c2 ~z) = 0
The trickiest part here is writing everything down in terms of dot products!
As soon as that’s done, it’s very easy. Using vector identities.
~x · (c1 ~y + c2 ~z) = ~x · (c1 ~y ) + ~x · (c2 ~z)
= c1 ~x · ~y + c2 ~x · ~z = 0
and so we’re done.
6
15. Calculate proj~a~b in each case, and verify that ~b − proj~a~b is orthogonal to
~a.
(b) ~a = [−5, 3, 0], ~b = [3, −7, 1].
Solution: By definition,
proj~a~b =
~a · ~b
k~ak2
!
~a
Therefore, here we have
[−5, 3, 0] · [3, −7, 1]
~
[−5, 3, 0]
proj~a b =
k[−5, 3, 0]k2
−36
=
[−5, 3, 0]
(−5)2 + 32 + 02
36
90 54
= − [−5, 3, 0] =
,− ,0
34
17 17
Now, verifying that ~b − proj~a~b is orthogonal to ~a:
90 54
~
~
~a · b − proj~a b = [−5, 3, 0] · [3, −7, 1] −
,− ,0
17 17
39 65
= [−5, 3, 0] · − , − , 1
17 17
195 195
=
−
=0
17
17
Since the dot product is 0, they are indeed orthogonal.
23. True or False:
(a) For any vectors ~x and ~y , and any scalar d, ~x · (d~y ) = (d~x) · ~y .
TRUE: This follows from Identity (4) in Theorem 1.5.
(b) For all ~x, ~y in Rn with ~x 6= ~0, ~x · ~y /k~xk ≤ k~y k.
TRUE: This follows from Cauchy-Schwarz.
(c) For all ~x, ~y in Rn , k~x − ~y k ≤ k~xk − k~y k.
FALSE: In fact, the opposite of this is true, which can be shown
using Triangle Inequality: k~x − ~y k ≥ k~xk − k~y k. Let’s provide a
counterexample: let ~x = [1, 0] and ~y = [0, 1]. In that case,
√
k~x − ~y k = k[1, 1]k = 2, k~xk = 1, k~y k = 1
7
and it’s not true that
√
2 ≤ 1 − 1 = 0.
(d) If θ is the angle between ~x and ~y in Rn , and θ >
π
2,
then ~x · ~y > 0.
FALSE: From Theorem 1.8, ~x · ~y > 0 if and only if the angle θ
is acute.
(e) The standard unit vectors in Rn are mutually orthogonal.
TRUE: The standard unit vectors are ~e1 = [1, 0, . . . , 0], ~e2 = [0, 1, 0, . . . ],
etc., and it’s easy to check that any pair of these has dot product 0.
(f) If proj~a~b = ~b, then ~a is perpendicular to ~b.
FALSE: It can be checked that ~a is perpendicular to ~b if and only
if proj~a~b = ~0. Furthermore, proj~a~b = ~b if and only if ~a and ~b are
parallel. We will not prove these statements here, but we will provide a counterexample to the assertion. Take ~a = ~b = [1, 0]. In that
case, it can be checked that proj~a~b = ~b, but ~a and ~b are certainly not
orthogonal.
8