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Unformatted text preview: LINEAR ALGEBRA
W W L CHEN
c W W L Chen, 1982, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990.
It is available free to all individuals, on the understanding that it is not to be used for ﬁnancial gain,
and may be downloaded and/or photocopied, with or without permission from the author.
However, this document may not be kept on any information storage and retrieval system without permission
from the author, unless such system is not accessible to any individuals other than its owners. Chapter 4
VECTORS 4.1. Introduction
A vector is an object which has magnitude and direction.
Example 4.1.1. We may be travelling northeast at 50 kph. In this case, the direction of the velocity
is northeast and the magnitude of the velocity is 50 kph. We can describe our velocity in kph as
50 50
√ ,√
2
2 , where the ﬁrst coordinate describes the speed with which we are moving east and the second coordinate
describes the speed with which we are moving north.
Example 4.1.2. An object in the sky may be 100 metres away in the southeast direction 45 degrees
upwards. In this case, the direction of its position is southeastand 45 degrees upwards and the magnitude
of its distance is 100 metres. We can describe the position of the object in metres as
100
50, −50, √
2 , where the ﬁrst coordinate describes the distance east, the second coordinate describes the distance north
and the third coordinate describes the distance up.
The purpose of this chapter is to study some relationship between algebra and geometry. We shall
ﬁrst study some algebra which is motivated by geometric considerations. We then use the algebra later
to better understand some problems in geometry.
Chapter 4 : Vectors page 1 of 17 c Linear Algebra W W L Chen, 1982, 2005 4.2. Vectors in R2
A vector on the plane R2 can be described as an ordered pair u = (u1 , u2 ), where u1 , u2 ∈ R.
Definition. Two vectors u = (u1 , u2 ) and v = (v1 , v2 ) in R2 are said to be equal, denoted by u = v, if
u1 = v1 and u2 = v2 .
Definition. For any two vectors u = (u1 , u2 ) and v = (v1 , v2 ) in R2 , we deﬁne their sum to be
u + v = (u1 , u2 ) + (v1 , v2 ) = (u1 + v1 , u2 + v2 ).
−
−
→
−→
−
Geometrically, if we represent the two vectors u and v by AB and BC respectively, then the sum
−
→
u + v is represented by AC as shown in the diagram below: A I
r8 C
rr
rr
rr
rr
rr
r
u+v rrr
v
r
rr
rr
rr
rr
rr
rr
rr
/
B u The next diagram demonstrates geometrically that u + v = v + u: A / u 8C
qq I
qq
qq
qq
qq
q
qq
u+vqqq
v
v
qq
qq
qq
qq
qq
qq
qq
qq
/
D
I u B PROPOSITION 4A. (VECTOR ADDITION)
(a) For every u, v ∈ R2 , we have u + v ∈ R2 .
(b) For every u, v, w ∈ R2 , we have u + (v + w) = (u + v) + w.
(c) For every u ∈ R2 , we have u + 0 = u, where 0 = (0, 0) ∈ R2 .
(d) For every u ∈ R2 , there exists v ∈ R2 such that u + v = 0.
(e) For every u, v ∈ R2 , we have u + v = v + u.
Proof. Write u = (u1 , u2 ), v = (v1 , v2 ) and w = (w1 , w2 ), where u1 , u2 , v1 , v2 , w1 , w2 ∈ R. To check
part (a), simply note that u1 + v1 , u2 + v2 ∈ R. To check part (b), note that
u + (v + w) = (u1 , u2 ) + (v1 + w1 , v2 + w2 ) = (u1 + (v1 + w1 ), u2 + (v2 + w2 ))
= ((u1 + v1 ) + w1 , (u2 + v2 ) + w2 ) = (u1 + v1 , u2 + v2 ) + (w1 , w2 )
= (u + v) + w.
Part (c) is trivial. Next, if v = (−u1 , −u2 ), then u + v = 0, giving part (d). To check part (e), note
that u + v = (u1 + v1 , u2 + v2 ) = (v1 + u1 , v2 + u2 ) = v + u.
Chapter 4 : Vectors page 2 of 17 c Linear Algebra W W L Chen, 1982, 2005 Definition. For any vector u = (u1 , u2 ) in R2 and any scalar c ∈ R, we deﬁne the scalar multiple to be
cu = c(u1 , u2 ) = (cu1 , cu2 ).
Example 4.2.1. Suppose that u = (2, 1). Then −2u = (−4, 2). Geometrically, if we represent the two
−
→
−→
−
vectors u and −2u by OA and OB respectively, then we have the diagram below: B O
pp
pp
p
pp
pp
pp
pp
pp −2u
pp
pp
pp
wpp pp
pp
pp
p
pp u
pp 8A PROPOSITION 4B. (SCALAR MULTIPLICATION)
(a) For every c ∈ R and u ∈ R2 , we have cu ∈ R2 .
(b) For every c ∈ R and u, v ∈ R2 , we have c(u + v) = cu + cv.
(c) For every a, b ∈ R and u ∈ R2 , we have (a + b)u = au + bu.
(d) For every a, b ∈ R and u ∈ R2 , we have (ab)u = a(bu).
(e) For every u ∈ R2 , we have 1u = u.
Proof. Write u = (u1 , u2 ) and v = (v1 , v2 ), where u1 , u2 , v1 , v2 ∈ R. To check part (a), simply note
that cu1 , cu2 ∈ R. To check part (b), note that
c(u + v) = c(u1 + v1 , u2 + v2 ) = (c(u1 + v1 ), c(u2 + v2 ))
= (cu1 + cv1 , cu2 + cv2 ) = (cu1 , cu2 ) + (cv1 , cv2 ) = cu + cv.
To check part (c), note that
(a + b)u = ((a + b)u1 , (a + b)u2 ) = (au1 + bu1 , au2 + bu2 )
= (au1 , au2 ) + (bu1 , bu2 ) = au + bu.
To check part (d), note that
(ab)u = ((ab)u1 , (ab)u2 ) = (a(bu1 ), a(bu2 )) = a(bu1 , bu2 ) = a(bu).
Finally, to check part (e), note that 1u = (1u1 , 1u2 ) = (u1 , u2 ) = u.
Definition. For any vector u = (u1 , u2 ) in R2 , we deﬁne the norm of u to be the nonnegative real
number
u= u2 + u2 .
1
2 Remarks. (1) The norm of a vector is simply its magnitude or length. The deﬁnition follows from the
famous theorem of Pythagoras.
(2) Suppose that P (u1 , u2 ) and Q(v1 , v2 ) are two points on the plane R2 . To calculate the distance
d(P, Q) between the two points, we can ﬁrst ﬁnd a vector from P to Q. This is given by (v1 − u1 , v2 − u2 ).
The distance d(P, Q) is then the norm of this vector, so that
d(P, Q) =
Chapter 4 : Vectors (v1 − u1 )2 + (v2 − u2 )2 .
page 3 of 17 c Linear Algebra W W L Chen, 1982, 2005 (3) It is not diﬃcult to see that for any vector u ∈ R2 and any scalar c ∈ R, we have cu = c u .
Definition. Any vector u ∈ R2 satisfying u = 1 is called a unit vector.
Example 4.2.2. The vector (3, 4) has norm 5.
Example 4.2.3. The distance between the points (6, 3) and (9, 7) is (9 − 6)2 + (7 − 3)2 = 5. Example 4.2.4. The vectors (1, 0) and (0, −1) are unit vectors in R2 .
√
√
Example 4.2.5. The unit vector in the direction of the vector (1, 1) is (1/ 2, 1/ 2).
Example 4.2.6. In fact, all unit vectors in R2 are of the form (cos θ, sin θ), where θ ∈ R.
Quite often, we may want to ﬁnd the angle between two vectors. The scalar product of the two vectors
then comes in handy. We shall deﬁne the scalar product in two ways, one in terms of the angle between
the two vectors and the other not in terms of this angle, and show that the two deﬁnitions are in fact
equivalent.
Definition. Suppose that u = (u1 , u2 ) and v = (v1 , v2 ) are vectors in R2 , and that θ ∈ [0, π ] represents
the angle between them. We deﬁne the scalar product u · v of u and v by
(1) u v cos θ u·v = if u = 0 and v = 0,
if u = 0 or v = 0. 0 Alternatively, we write
u · v = u1 v1 + u2 v2 . (2) The deﬁnitions (1) and (2) are clearly equivalent if u = 0 or v = 0. On the other hand, we have the
following result.
PROPOSITION 4C. Suppose that u = (u1 , u2 ) and v = (v1 , v2 ) are nonzero vectors in R2 , and that
θ ∈ [0, π ] represents the angle between them. Then
u v cos θ = u1 v1 + u2 v2 .
−
→
−→
−
Proof. Geometrically, if we represent the two vectors u and v by OA and OB respectively, then the
−
−
→
diﬀerence v − u is represented by AB as shown in the diagram below:
B1
I X1 1
1
1
1 v−u
1
1
1
1
1 v O 7A
oo
oo
oo
o
oo u
θ ooo
o By the Law of cosines, we have
2 2 2 AB = OA + OB − 2OA OB cos θ;
Chapter 4 : Vectors page 4 of 17 c Linear Algebra W W L Chen, 1982, 2005 in other words, we have
v−u 2 =u 2 +v 2 − 2 u v cos θ, so that
u v cos θ = 1 ( u
2 2 +v 2 − v − u 2) 2
2
= 1 (u2 + u2 + v1 + v2 − (v1 − u1 )2 − (v2 − u2 )2 )
1
2
2
= u1 v1 + u2 v2 as required.
Remarks. (1) We say that two nonzero vectors in R2 are orthogonal if the angle between them is π/2.
It follows immediately from the deﬁnition of the scalar product that two nonzero vectors u, v ∈ R2 are
orthogonal if and only if u · v = 0.
(2) We can calculate the scalar product of any two nonzero vectors u, v ∈ R2 by the formula (2) and
then use the formula (1) to calculate the angle between u and v.
√
√
Example 4.2.7. Suppose that u = ( 3, 1) and v = ( 3, 3). Then by the formula (2), we have
u · v = 3 + 3 = 6.
Note now that
u =2 and √
v = 2 3. It follows from the formula (1) that
√
u·v
3
6
cos θ =
=√=
,
uv
2
43
so that θ = π/6.
√
√
Example 4.2.8. Suppose that u = ( 3, 1) and v = (− 3, 3). Then by the formula (2), we have
u · v = 0. It follows that u and v are orthogonal.
PROPOSITION 4D. (SCALAR PRODUCT) Suppose that u, v, w ∈ R2 and c ∈ R. Then
(a) u · v = v · u;
(b) u · (v + w) = (u · v) + (u · w);
(c) c(u · v) = (cu) · v = u · (cv);
(d) u · u ≥ 0; and
(e) u · u = 0 if and only if u = 0.
Proof. Write u = (u1 , u2 ), v = (v1 , v2 ) and w = (w1 , w2 ), where u1 , u2 , v1 , v2 , w1 , w2 ∈ R. Part (a) is
trivial. To check part (b), note that
u · (v + w) = u1 (v1 + w1 ) + u2 (v2 + w2 ) = (u1 v1 + u2 v2 ) + (u1 w1 + u2 w2 ) = u · v + u · w.
Part (c) is rather simple. To check parts (d) and (e), note that u · u = u2 + u2 ≥ 0, and that equality
1
2
holds precisely when u1 = u2 = 0.
Chapter 4 : Vectors page 5 of 17 c Linear Algebra W W L Chen, 1982, 2005 Consider the diagram below: o o o o o o R/
W/
(3) /
/
/
/
/
v/
/
/
/
/
/ O u o P/
J / / / / / / 7Q
oo
oo
oo
o
oo w
oo
oo q
a q q q q8 A −
→
−
−
→
Here we represent the two vectors a and u by OA and OP respectively. If we project the vector u on to
−→
−
the line OA, then the image of the projection is the vector w, represented by OQ. On the other hand,
if we project the vector u on to a line perpendicular to the line OA, then the image of the projection is
−
−
→
the vector v, represented by OR.
Definition. In the notation of the diagram (3), the vector w is called the orthogonal projection of the
vector u on the vector a, and denoted by w = proja u.
PROPOSITION 4E. (ORTHOGONAL PROJECTION) Suppose that u, a ∈ R2 . Then
u·a
proja u =
a.
a2
−
−
→
Remark. Note that the component of u orthogonal to a, represented by OR in the diagram (3), is
u·a
u − proja u = u −
a.
a2
Proof of Proposition 4E. Note that w = k a for some k ∈ R. It clearly suﬃces to prove that
u·a
k=
.
a2
It is easy to see that the vectors u − w and a are orthogonal. It follows that the scalar product
(u − w) · a = 0. In other words, (u − k a) · a = 0. Hence
u·a
u·a
k=
=
a·a
a2
as required.
To end this section, we shall apply our knowledge gained so far to ﬁnd a formula that gives the
perpendicular distance of a point (x0 , y0 ) from a line ax + by + c = 0. Consider the diagram below:
8P
pp J /
pp
/
Dpp
p
/
pp
 pp
 xpp
/
n6
ax+by +c=0 /
n
nn
/
nn
u
n
n
/
nn
nn
nn
oQ
oo
oo
o
oo
oo
oo
o = (a, b) O Chapter 4 : Vectors page 6 of 17 c Linear Algebra W W L Chen, 1982, 2005 Suppose that (x1 , y1 ) is any arbitrary point O on the line ax + by + c = 0. For any other point (x, y ) on
the line ax + by + c = 0, the vector (x − x1 , y − y1 ) is parallel to the line. On the other hand,
(a, b) · (x − x1 , y − y1 ) = (ax + by ) − (ax1 + by1 ) = −c + c = 0,
−→
−
so that the vector n = (a, b), in the direction OQ, is perpendicular to the line ax + by + c = 0.
Suppose next that the point (x0 , y0 ) is represented by the point P in the diagram. Then the vector
−
−
→
−→
−
u = (x0 − x1 , y0 − y1 ) is represented by OP , and OQ represents the orthogonal projection projn u of u
on the vector n. Clearly the perpendicular distance D of the point (x0 , y0 ) from the line ax + by + c = 0
satisﬁes
D = projn u = u·n
(x0 − x1 , y0 − y1 ) · (a, b)
ax0 + by0 − ax1 − by1 
ax0 + by0 + c
√
√
√
n=
=
=
.
2 + b2
2 + b2
n2
a
a
a2 + b2 We have proved the following result.
PROPOSITION 4F. The perpendicular distance D of a point (x0 , y0 ) from a line ax + by + c = 0 is
given by
D= ax0 + by0 + c
√
.
a2 + b2 Example 4.2.9. The perpendicular distance D of the point (5, 7) from the line 2x − 3y + 5 = 0 is given
by
D= 10 − 21 + 5
6
√
=√ .
4+9
13 4.3. Vectors in R3
In this section, we consider the same problems as in Section 4.2, but in 3space R3 . Any reader who
feels conﬁdent may skip this section.
A vector on the plane R3 can be described as an ordered triple u = (u1 , u2 , u3 ), where u1 , u2 , u3 ∈ R.
Definition. Two vectors u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) in R3 are said to be equal, denoted by
u = v, if u1 = v1 , u2 = v2 and u3 = v3 .
Definition. For any two vectors u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) in R3 , we deﬁne their sum to be
u + v = (u1 , u2 , u3 ) + (v1 , v2 , v3 ) = (u1 + v1 , u2 + v2 , u3 + v3 ).
Definition. For any vector u = (u1 , u2 , u3 ) in R3 and any scalar c ∈ R, we deﬁne the scalar multiple
to be
cu = c(u1 , u2 , u3 ) = (cu1 , cu2 , cu3 ). The following two results are the analogues of Propositions 4A and 4B. The proofs are essentially
similar.
Chapter 4 : Vectors page 7 of 17 c Linear Algebra W W L Chen, 1982, 2005 PROPOSITION 4A’. (VECTOR ADDITION)
(a) For every u, v ∈ R3 , we have u + v ∈ R3 .
(b) For every u, v, w ∈ R3 , we have u + (v + w) = (u + v) + w.
(c) For every u ∈ R3 , we have u + 0 = u, where 0 = (0, 0, 0) ∈ R3 .
(d) For every u ∈ R3 , there exists v ∈ R3 such that u + v = 0.
(e) For every u, v ∈ R3 , we have u + v = v + u.
PROPOSITION 4B’. (SCALAR MULTIPLICATION)
(a) For every c ∈ R and u ∈ R3 , we have cu ∈ R3 .
(b) For every c ∈ R and u, v ∈ R3 , we have c(u + v) = cu + cv.
(c) For every a, b ∈ R and u ∈ R3 , we have (a + b)u = au + bu.
(d) For every a, b ∈ R and u ∈ R3 , we have (ab)u = a(bu).
(e) For every u ∈ R3 , we have 1u = u.
Definition. For any vector u = (u1 , u2 , u3 ) in R3 , we deﬁne the norm of u to be the nonnegative real
number
u= u2 + u2 + u2 .
1
2
3 Remarks. (1) Suppose that P (u1 , u2 , u3 ) and Q(v1 , v2 , v3 ) are two points in R3 . To calculate the
distance d(P, Q) between the two points, we can ﬁrst ﬁnd a vector from P to Q. This is given by
(v1 − u1 , v2 − u2 , v3 − u3 ). The distance d(P, Q) is then the norm of this vector, so that
d(P, Q) = (v1 − u1 )2 + (v2 − u2 )2 + (v3 − u3 )2 . (2) It is not diﬃcult to see that for any vector u ∈ R3 and any scalar c ∈ R, we have
cu = c u .
Definition. Any vector u ∈ R3 satisfying u = 1 is called a unit vector.
Example 4.3.1. The vector (3, 4, 12) has norm 13.
Example 4.3.2. The distance between the points (6, 3, 12) and (9, 7, 0) is 13.
Example 4.3.3. The vectors (1, 0, 0) and (0, −1, 0) are unit vectors in R3 .
√
√
Example 4.3.4. The unit vector in the direction of the vector (1, 0, 1) is (1/ 2, 0, 1/ 2).
The theory of scalar products can be extended to R3 is the natural way.
Definition. Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are vectors in R3 , and that θ ∈ [0, π ]
represents the angle between them. We deﬁne the scalar product u · v of u and v by
(4) u·v = u v cos θ
0 if u = 0 and v = 0,
if u = 0 or v = 0. Alternatively, we write
(5) u · v = u1 v1 + u2 v2 + u3 v3 . The deﬁnitions (4) and (5) are clearly equivalent if u = 0 or v = 0. On the other hand, we have the
following analogue of Proposition 4C. The proof is similar.
Chapter 4 : Vectors page 8 of 17 c Linear Algebra W W L Chen, 1982, 2005 PROPOSITION 4C’. Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are nonzero vectors in R3 ,
and that θ ∈ [0, π ] represents the angle between them. Then
u v cos θ = u1 v1 + u2 v2 + u3 v3 .
Remarks. (1) We say that two nonzero vectors in R3 are orthogonal if the angle between them is π/2.
It follows immediately from the deﬁnition of the scalar productthat two nonzero vectors u, v ∈ R3 are
orthogonal if and only if u · v = 0.
(2) We can calculate the scalar product of any two nonzero vectors u, v ∈ R3 by the formula (5) and
then use the formula (4) to calculate the angle between u and v.
√
Example 4.3.5. Suppose that u = (2, 0, 0) and v = (1, 1, 2). Then by the formula (5), we have
u · v = 2. Note now that u = 2 and v = 2. It follows from the formula (4) that
cos θ = u·v
2
1
==,
uv
4
2 so that θ = π/3.
Example 4.3.6. Suppose that u = (2, 3, 5) and v = (1, 1, −1). Then by the formula (5), we have
u · v = 0. It follows that u and v are orthogonal.
The following result is the analogue of Proposition 4D. The proof is similar.
PROPOSITION 4D’. (SCALAR PRODUCT) Suppose that u, v, w ∈ R3 and c ∈ R. Then
(a) u · v = v · u;
(b) u · (v + w) = (u · v) + (u · w);
(c) c(u · v) = (cu) · v = u · (cv);
(d) u · u ≥ 0; and
(e) u · u = 0 if and only if u = 0.
Suppose now that a and u are two vectors in R3 . Then since two vectors are always coplanar, we can
draw the following diagram which represents the plane they lie on: o o o o o o R/
W/
(6) /
/
/
/
/
v/
/
/
/
/
/ O u o P/
J / / / / / / o7 Q
oo
oo
o
oo
oo w
oo
o q
a q q q q8 A Note that this diagram is essentially the same as the diagram (3), the only diﬀerence being that while
the diagram (3) shows the whole of R2 , the diagram (6) only shows part of R3 . As before, we represent
−
→
−
−
→
the two vectors a and u by OA and OP respectively. If we project the vector u on to the line OA, then
−→
−
the image of the projection is the vector w, represented by OQ. On the other hand, if we project the
vector u on to a line perpendicular to the line OA, then the image of the projection is the vector v,
−
−
→
represented by OR.
Definition. In the notation of the diagram (6), the vector w is called the orthogonal projection of the
vector u on the vector a, and denoted by w = proja u.
Chapter 4 : Vectors page 9 of 17 c Linear Algebra W W L Chen, 1982, 2005 The following result is the analogue of Proposition 4E. The proof is similar.
PROPOSITION 4E’. (ORTHOGONAL PROJECTION) Suppose that u, a ∈ R3 . Then
proja u = u·a
a.
a2 −
−
→
Remark. Note that the component of u orthogonal to a, represented by OR in the diagram (6), is
u − proja u = u − u·a
a.
a2 4.4. Vector Products
In this section, we shall discuss a product of vectors unique to R3 . The idea of vector products has wide
applications in geometry, physics and engineering, and is motivated by the wish to ﬁnd a vector that is
perpendicular to two given vectors.
We shall use the right hand rule. In other words, if we hold the thumb on the right hand upwards
and close the remaining four ﬁngers, then the ﬁngers point from the xdirection towards the y direction,
while the thumb points towards the z direction. Alternatively, if we imagine Columbus had never lived
and that the earth were ﬂat, then taking the xdirection as east and the y direction as north, then the
z direction is upwards!
We shall frequently use the three vectors i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1) in R3 .
Definition. Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are two vectors in R3 . Then the vector
product u × v is deﬁned by the determinant i
u × v = det u1
v1 k
u3 .
v3 j
u2
v2 Remarks. (1) Note that
i × j = −(j × i) = k,
j × k = −(k × j) = i,
k × i = −(i × k) = j.
(2) Using cofactor expansion by row 1, we have
u × v = det
= det u2
v2
u2
v2 u3
v3
u3
v3 i − det
, − det u1
v1
u1
v1 u3
v3
u3
v3 j + det
, det u1
v1
u1
v1 u2
v2 k u2
v2 = (u2 v3 − u3 v2 , u3 v1 − u1 v3 , u1 v2 − u2 v1 ).
We shall ﬁrst of all show that the vector product u × v is orthogonal to both u and v.
Chapter 4 : Vectors page 10 of 17 c Linear Algebra W W L Chen, 1982, 2005 PROPOSITION 4G. Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are two vectors in R3 . Then
(a) u · (u × v) = 0; and
(b) v · (u × v) = 0.
Proof. Note ﬁrst of all that
u · (u × v) = (u1 , u2 , u3 ) · det
= u1 det u2
v2 u3
v3 u2
v2 u3
v3 − u2 det , − det
u1
v1 u3
v3 in view of cofactor expansion by row 1. On the other u1 u2
det u1 u2
v1 v2 u1
v1 u3
v3 + u3 det , det
u1
v1 u2
v2 u1
v1 u1
= det u1
v1 u2
v2 u3
u3 ,
v3 u2
u2
v2 hand, clearly u3
u3 = 0.
v3 This proves part (a). The proof of part (b) is similar.
Example 4.4.1. Suppose that u = (1, −1, 2) and v = (3, 0, 2). Then i
jk
−1 2
12
u × v = det 1 −1 2 = det
, − det
, det
02
32
302 1
3 −1
0 = (−2, 4, 3). Note that (1, −1, 2) · (−2, 4, 3) = 0 and (3, 0, 2) · (−2, 4, 3) = 0.
PROPOSITION 4H. (VECTOR PRODUCT) Suppose that u, v, w ∈ R3 and c ∈ R. Then
(a) u × v = −(v × u);
(b) u × (v + w) = (u × v) + (u × w);
(c) (u + v) × w = (u × w) + (v × w);
(d) c(u × v) = (cu) × v = u × (cv);
(e) u × 0 = 0; and
(f ) u × u = 0.
Proof. Write u = (u1 , u2 , u3 ), v = (v1 , v2 , v3 ) and w = (w1 , w2 , w3 ). To check part (a), note that i
j
k
i
j
k
det u1 u2 u3 = − det v1 v2 v3 .
v1 v2 v3
u1 u2 u3
To check part (b), note that i
j
det u1
u2
v1 + w1 v2 + w2 k
i
u3 = det u1
v3 + w3
v1 Part (c) is similar. To check part (d), note that i
j
k
i
c det u1 u2 u3 = det cu1
v1 v2 v3
v1
To check parts (e) and (f), note that i
j
k
u × 0 = det u1 u2 u3 = 0
0
0
0 j
cu2
v2 j
u2
v2 k
i
u3 + det u1
v3
w1 k
i
cu3 = det u1
v3
cv1 and i
u × u = det u1
u1 j
u2
cv2 j
u2
u2 j
u2
w2 k
u3 .
w3 k
u3 .
cv3 k
u3 = 0
u3 as required.
Chapter 4 : Vectors page 11 of 17 c Linear Algebra W W L Chen, 1982, 2005 Next, we shall discuss an application of vector product to the evaluaton of the area of a parallelogram.
To do this, we shall ﬁrst establish the following result.
PROPOSITION 4J. Suppose that u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) are nonzero vectors in R3 ,
and that θ ∈ [0, π ] represents the angle between them. Then
(a) u × v 2 = u 2 v 2 − (u · v)2 ; and
(b) u × v = u v sin θ.
Proof. Note that
u×v (7) 2 = (u2 v3 − u3 v2 )2 + (u3 v1 − u1 v3 )2 + (u1 v2 − u2 v1 )2 and
(8) u 2 v 2 2
2
2
− (u · v)2 = (u2 + u2 + u2 )(v1 + v2 + v3 ) − (u1 v1 + u2 v2 + u3 v3 )2 .
1
2
3 Part (a) follows on expanding the right hand sides of (7) and (8) and checking that they are equal. To
prove part (b), recall that
u · v = u v cos θ.
Combining with part (a), we obtain
u×v 2 =u 2 v 2 −u 2 v 2 cos2 θ = u 2 v 2 sin2 θ. Part (b) follows.
−
→
Consider now a parallelogram with vertices O, A, B, C . Suppose that u and v are represented by OA
−
−
→
and OC respectively. If we imagine the side OA to represent the base of the parallelogram, so that
the base has length u , then the height of the the parallelogram is given by v sin θ, as shown in the
diagram below:
O
C
B
I v v sin θ O
o θ
u u /A
/ It follows from Proposition 4J that the area of the parallelogram is given by u × v . We have proved
the following result.
PROPOSITION 4K. Suppose that u, v ∈ R3 . Then the parallelogram with u and v as two of its sides
has area u × v .
We conclude this section by making a remark on the vector product u × v of two vectors in R3 .
Recall that the vector product is perpendicular to both u and v. Furthermore, it can be shown that the
direction of u × v satisﬁes the right hand rule, in the sense that if we hold the thumb on the right hand
outwards and close the remaining four ﬁngers, then the thumb points towards the u × vdirection when
the ﬁngers point from the udirection towards the vdirection. Also, we showed in Proposition 4J that
the magnitude of u × v depends only on the norm of u and v and the angle between the two vectors. It
Chapter 4 : Vectors page 12 of 17 c Linear Algebra W W L Chen, 1982, 2005 follows that the vector product is unchanged as long as we keep a right hand coordinate system. This is
an important consideration in physics and engineering, where we may use diﬀerent coordinate systems
on the same problem. 4.5. Scalar Triple Products
Suppose that u, v, w ∈ R3 do not all lie on the same plane. Consider the parallelepiped with u, v, w as
three of its edges. We are interested in calculating the volume of this parallelepiped. Suppose that u, v
− −→
→−
−
−
→
and w are represented by OA, OB and OC respectively. Consider the diagram below:
v×w O P _ _ _A
G mm; ;
mm ; ; ; ;
mm
;
mm
;;;
mm
;;; ; ;
mm
m
; ;;
;;; ;;; qq
qq
qq
q
qq
qq ; ;;
; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; q;;;
u
o7 C
q
oo
qq
oo
qq
oo
oo w
qq
oo
qq
oo
/Bq
O
v
;;; By Proposition 4K, the base of this parallelepiped, with O, B, C as three of the vertices, has area v × w .
−
−
→
Next, note that if OP is perpendicular to the base of the parallelepiped, then OP is in the direction of
v × w. If P A is perpendicular to OP , then the height of the parallelepiped is equal to the norm of the
orthogonal projection of u on v × w. In other words, the parallelepiped has height
projv×w u = u · (v × w)
u · (v × w)
(v × w) =
.
v×w 2
v×w Hence the volume of the parallelepiped is given by
V = u · (v × w).
We have proved the following result.
PROPOSITION 4L. Suppose that u, v, w ∈ R3 . Then the parallelepiped with u, v and w as three of
its edges has volume u · (v × w).
Definition. Suppose that u, v, w ∈ R3 . Then u · (v × w) is called the scalar triple product of u, v and
w.
Remarks. (1) It follows immediately from Proposition 4L that three vectors in R3 are coplanar if and
only if their scalar triple product is zero.
(2) Note that
(9) u · (v × w) = (u1 , u2 , u3 ) · det
= u1 det v2
w2 u1
= det v1
w1 v3
w3
u2
v2
w2 v2
w2 v3
w3 − u2 det , − det
v1
w1 v3
w3 v1
w1 v3
w3 + u3 det , det v1
w1 v1
w1 v2
w2 v2
w2 u3
v3 ,
w3 in view of cofactor expansion by row 1.
Chapter 4 : Vectors page 13 of 17 c Linear Algebra W W L Chen, 1982, 2005 (3) It follows from identity (9) that
u · (v × w) = v · (w × u) = w · (u × v).
Note that each of the determinants can be obtained from the other two by twice interchanging two rows.
Example 4.5.1. Suppose that u = (1, 0, 1), v = (2, 1, 3) 1
u · (v × w) = det 2
0 and w = (0, 1, 1). Then 01
1 3 = 0,
11 so that u, v and w are coplanar.
Example 4.5.2. The volume of the parallelepiped with
three of its edges is given by 10
u · (v × w) = det 2 1
01 u = (1, 0, 1), v = (2, 1, 4) and w = (0, 1, 1) as 1
4 =  − 1 = 1.
1 4.6. Applications to Geometry in R3
In this section, we shall study lines and planes in R3 by using our results on vectors in R3 .
Consider ﬁrst of all a plane in R3 . Suppose that (x1 , y1 , z1 ) ∈ R3 is a given point on this plane.
Suppose further that n = (a, b, c) is a vector perpendicular to this plane. Then for any arbitrary point
(x, y, z ) ∈ R3 on this plane, the vector
(x, y, z ) − (x1 , y1 , z1 ) = (x − x1 , y − y1 , z − z1 )
must be perpendicular to n = (a, b, c). It follows that the scalar product
(a, b, c) · (x − x1 , y − y1 , z − z1 ) = 0,
so that
(10) a(x − x1 ) + b(y − y1 ) + c(z − z1 ) = 0. If we write −d = ax1 + by1 + cz1 , then (10) can be rewritten in the form
(11) ax + by + cz + d = 0. Equation (10) is usually called the pointnormal form of the equation of a plane, while equation (11) is
usually known as the general form of the equation of a plane.
Example 4.6.1. Consider the plane through the point (2, −5, 7) and perpendicular to the vector
(3, 5, −4). Here (a, b, c) = (3, 5, −4) and (x1 , y1 , z1 ) = (2, −5, 7). The equation of the plane is given
in pointnormal form by 3(x − 2) + 5(y + 5) − 4(z − 7) = 0, and in general form by 3x + 5y − 4z + 37 = 0.
Here −d = 6 − 25 − 28 = −37.
Consider next a line in R3 . Suppose that (x1 , y1 , z1 ) ∈ R3 is a given point on this line. Suppose further
that n = (a, b, c) is a vector parallel to this line. Then for any arbitrary point (x, y, z ) ∈ R3 on this line,
the vector
(x, y, z ) − (x1 , y1 , z1 ) = (x − x1 , y − y1 , z − z1 )
Chapter 4 : Vectors page 14 of 17 c Linear Algebra W W L Chen, 1982, 2005 must be parallel to n = (a, b, c). It follows that there is some number λ ∈ R such that
(x − x1 , y − y1 , z − z1 ) = λ(a, b, c),
so that
x = x1 + aλ,
y = y1 + bλ, (12) z = z1 + cλ,
where λ is called a parameter. Suppose further that a, b, c are all nonzero. Then, eliminating the
parameter λ, we obtain
x − x1
y − y1
z − z1
(13)
=
=
.
a
b
c
Equations (12) are usually called the parametric form of the equations of a line, while equations (13)
are usually known as the symmetric form of the equations of a line.
Example 4.6.2. Consider the line through the point (2, −5, 7) and parallel to the vector (3, 5, −4). Here
(a, b, c) = (3, 5, −4) and (x1 , y1 , z1 ) = (2, −5, 7). The equations of the line are given in parametric form
by
x = 2 + 3λ,
y = −5 + 5λ,
z = 7 − 4λ,
and in symmetric form by
x−2
y+5
z−7
=
=−
.
3
5
4
We next turn our attention to the question of ﬁnding the distance of a point from a plane. We shall
prove the following analogue of Proposition 4F.
PROPOSITION 4F’. The perpendicular distance D of a point (x0 , y0 , z0 ) from a plane ax+by +cz +d =
0 is given by
D= ax0 + by0 + cz0 + d
√
.
a2 + b2 + c2 Proof. Consider the following diagram:
;;
; ; ; ;;;;;
;;; ; ;
;;;
;;;
;;
;;;
;;; ; ;
; ; ; 8 P/
;;; ;
pJ
;
pp;;;;
/
;;; ; ;
Dppp ; ; ;
;;;
;
pp
/
;;;
;;; ; ;
pp
;;;
;
xp
pp
;;; /
n 6 = (a, b, c)
;;
/
;;;
m
;;
;;;
mm
mm
u
;;
;;; /
mm
;;;
/
mm
;;
mm
;;;
;;
; ; ; Q mm
;;
;;
oo
oo ; ; ; ;
;;
;;;
oo
o
;;
;;;
oo
;;
oo
;;;
oo
ax+by +cz =0 ;
O
;;
;;;; ;
;
;;
;;;;
;;
;;;; ; ; ;
;;
;;
;;
;;;;
;;
;;;; ; ; ;
;; ;;;;
;
Chapter 4 : Vectors page 15 of 17 c Linear Algebra W W L Chen, 1982, 2005 Suppose that (x1 , x2 , x3 ) is any arbitrary point O on the plane ax + by + cz + d = 0. For any other point
(x, y, z ) on the plane ax + by + cz + d = 0, the vector (x − x1 , y − y1 , z − z1 ) is parallel to the plane. On
the other hand,
(a, b, c) · (x − x1 , y − y1 , z − z1 ) = (ax + by + cz ) − (ax1 + by1 + cz1 ) = −d + d = 0,
−→
−
so that the vector n = (a, b, c), in the direction OQ, is perpendicular to the plane ax + by + cz + d = 0.
Suppose next that the point (x0 , y0 , z0 ) is represented by the point P in the diagram. Then the vector
−
−
→
−→
−
u = (x0 − x1 , y0 − y1 , z0 − z1 ) is represented by OP , and OQ represents the orthogonal projection projn u
of u on the vector n. Clearly the perpendicular distance D of the point (x0 , y0 , z0 ) from the plane
ax + by + cz + d = 0 satisﬁes
u·n
(x0 − x1 , y0 − y1 , z0 − z1 ) · (a, b, c)
√
n=
2
n
a2 + b2 + c2
ax0 + by0 + cz0 − ax1 − by1 − cz1 
ax0 + by0 + cz0 + d
√
√
=
=
2 + b2 + c2
a
a2 + b2 + c2 D = projn u = as required.
A special case of Proposition 4F’ is when (x0 , y0 , z0 ) = (0, 0, 0) is the origin. This show that the
perpendicular distance of the plane ax + by + cz + d = 0 from the origin is
√ a2 d
.
+ b2 + c2 Example 4.6.3. Consider the plane 3x + 5y − 4z + 37 = 0. The distance of the point (1, 2, 3) from the
plane is
√
3 + 10 − 12 + 37
38
19 2
√
.
=√ =
5
9 + 25 + 16
50
The distance of the origin from the plane is
√ 37
37
=√ .
9 + 25 + 16
50 Example 4.6.4. Consider also the plane 3x +5y − 4z − 1 = 0. Note that this plane is also perpendicular to
the vector (3, 5, −4) and is therefore parallel to the plane 3x+5y −4z +37 = 0. It is therefore reasonable to
ﬁnd the perpendicular distance between these two parallel planes. Note that the perpendicular distance
between the two planes is equal to the perpendicular distance of any point on 3x + 5y − 4z − 1 = 0 from
the plane 3x + 5y − 4z + 37 = 0. Note now that (1, 2, 3) lies on the plane 3x + 5y − 4z − 1 = 0. It follows
√
from Example 4.6.3 that the distance between the two planes is 19 2/5. Problems for Chapter 4
1. For each of the following pairs of vectors in R2 , calculate u + 3v, u · v, u − v and ﬁnd the angle
between u and v:
a) u = (1, 1) and v = (−5, 0)
b) u = (1, 2) and v = (2, 1)
2. For each of the following pairs of vectors in R2 , calculate 2u − 5v, u − 2v , u · v and the angle
between u and v (to the nearest degree):
a) u = (1, 3) and v = (−2, 1)
b) u = (2, 0) and v = (−1, 2)
Chapter 4 : Vectors page 16 of 17 Linear Algebra c W W L Chen, 1982, 2005 3. For the two vectors u = (2, 3) and v = (5, 1) in the 2dimensional euclidean space R2 , determine
each of the following:
a) u − v
b) u
c) u · (u − v)
d) the angle between u and u − v
4. For each of the following pairs of vectors in R3 , calculate u + 3v, u · v, u − v , ﬁnd the angle
between u and v, and ﬁnd a unit vector perpendicular to both u and v:
a) u = (1, 1, 1) and v = (−5, 0, 5)
b) u = (1, 2, 3) and v = (3, 2, 1)
5. Find vectors v and w such that v is parallel to (1, 2, 3), v + w = (7, 3, 5) and w is orthogonal to
(1, 2, 3).
6. Consider the three points P (2, 3, 1), Q(4, 2, 5) and R(1, 6, −3).
a) Find the equation of the line through P and Q.
b) Find the equation of the plane perpendicular to the line in part (a) and passing through the point
R.
c) Find the distance between R and the line in part (a).
d) Find the area of the parallelogram with the three points as vertices.
e) Find the equation of the plane through the three points.
f) Find the distance of the origin (0, 0, 0) from the plane in part (e).
g) Are the planes in parts (b) and (e) perpendicular? Justify your assertion.
7. Consider the points (1, 2, 3), (0, 2, 4) and (2, 1, 3) in R3 .
a) Find the area of a parallelogram with these points as three of its vertices.
b) Find the perpendicular distance between the point (1, 2, 3) and the line passing through the points
(0, 2, 4) and (2, 1, 3).
8. Consider the points (1, 2, 3), (0, 2, 4) and (2, 1, 3) in R3 .
a) Find a vector perpendicular to the plane containing these points.
b) Find the equation of this plane and its perpendicular distance from the origin.
c) Find the equation of the line perpendicular to this plane and passing through the point (3, 6, 9).
9. Find the equation of the plane through the points (1, 2, −3), (2, −3, 4) and (−3, 1, 2).
10. Find the volume of a parallelepiped with the points (1, 2, 3), (0, 2, 4), (2, 1, 3) and (3, 6, 9) as four of
its vertices. Chapter 4 : Vectors page 17 of 17 ...
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This note was uploaded on 06/13/2009 for the course TAM 455 taught by Professor Petrina during the Fall '08 term at Cornell.
 Fall '08
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