W W L CHEN
c
W W L Chen, 1982, 2008.
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
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Chapter 4
VECTORS
4.1. Introduction
A vector is an object which has magnitude and direction.
Example 4.1.1.We may be travelling north-east at 50 kph. In this case, the direction of the velocity is north-east and the magnitude of the velocity is 50 kph. We can describe our velocity in kph as
50
√
2, 50
√
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 south-east direction 45 degrees upwards. In this case, the direction of its position is south-eastand 45 degrees upwards and the magnitude of its distance is 100 metres. We can describe the position of the object in metres as
50,−50,100√
2
,
where the first coordinate describes the distance east, the second coordinate describes the distance north and the third coordinate describes the distance up.
4.2. Vectors inR2
A vector on the planeR2 can be described as an ordered pairu= (u1, u2), whereu1, u2∈R.
Definition.Two vectorsu= (u1, u2) andv= (v1, v2) inR2are said to be equal, denoted byu=v, if
u1=v1 andu2=v2.
Definition.For any two vectorsu= (u1, u2) andv= (v1, v2) inR2, 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+vis represented by−→AC as shown in the diagram below:
C
A u B
u+v
v
The next diagram demonstrates geometrically thatu+v=v+u:
D C
A B
u
u u+v v
v
PROPOSITION 4A.(VECTOR ADDITION)
(a) For everyu,v∈R2, we haveu+v∈R2.
(b) For everyu,v,w∈R2, we haveu+ (v+w) = (u+v) +w.
(c) For everyu∈R2, we haveu+0=u, where0= (0,0)∈R2.
(d) For everyu∈R2, there existsv∈R2 such that u+v=0.
(e) For everyu,v∈R2, we haveu+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 thatu1+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.
Definition.For any vectoru= (u1, u2) inR2and any scalarc∈R, we define the scalar multiple to be
cu=c(u1, u2) = (cu1, cu2).
Example 4.2.1.Suppose thatu= (2,1). Then−2u= (−4,2). Geometrically, if we represent the two vectorsuand−2uby−→OAand−−→OB respectively, then we have the diagram below:
A
O
B
u
−2u
PROPOSITION 4B.(SCALAR MULTIPLICATION)
(a) For everyc∈Randu∈R2, we havecu∈R2.
(b) For everyc∈Randu,v∈R2, we have c(u+v) =cu+cv.
(c) For everya, b∈Randu∈R2, we have(a+b)u=au+bu.
(d) For everya, b∈Randu∈R2, we have(ab)u=a(bu).
(e) For everyu∈R2, we have1u=u.
Proof.Write u= (u1, u2) andv = (v1, v2), where u1, u2, v1, v2 ∈ R. To check part (a), simply note thatcu1, 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 uto be the non-negative real number
kuk=
q u2
1+u22.
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) andQ(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 fromP toQ. This is given by (v1−u1, v2−u2). The distanced(P, Q) is then the norm of this vector, so that
(3) It is not difficult to see that for any vectoru∈R2 and any scalarc∈R, we havekcuk=|c|kuk.
Definition.Any vectoru∈R2 satisfyingkuk= 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) isp(9−6)2+ (7−3)2= 5.
Example 4.2.4.The vectors (1,0) and (0,−1) are unit vectors inR2.
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 inR2 are of the form (cosθ,sinθ), whereθ∈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 thatu= (u1, u2) andv= (v1, v2) are vectors inR2, and thatθ∈[0, π] represents the angle between them. We define the scalar productu·vofuandvby
u·v=nkukkvkcosθ ifu6=0andv6=0,
0 ifu=0orv=0. (1)
Alternatively, we write
u·v=u1v1+u2v2. (2)
The definitions (1) and (2) are clearly equivalent ifu=0orv=0. On the other hand, we have the following result.
PROPOSITION 4C.Suppose thatu= (u1, u2)andv= (v1, v2)are non-zero vectors in R2, and that
θ∈[0, π] represents the angle between them. Then
kukkvkcosθ=u1v1+u2v2.
Proof.Geometrically, if we represent the two vectors uandv by −→OAand −−→OB respectively, then the differencev−uis represented by−−→ABas shown in the diagram below:
B
A
O
v−u
θ
u v
By the Law of cosines, we have
in other words, we have
kv−uk2=kuk2+kvk2−2kukkvkcosθ,
so that
kukkvkcosθ= 1 2(kuk
2+kvk2− kv−uk2)
= 1 2(u
2
1+u22+v21+v22−(v1−u1)2−(v2−u2)2)
=u1v1+u2v2
as required.
Remarks.(1) We say that two non-zero vectors inR2are orthogonal if the angle between them isπ/2.
It follows immediately from the definition of the scalar product that two non-zero vectorsu,v∈R2are orthogonal if and only ifu·v= 0.
(2) We can calculate the scalar product of any two non-zero vectorsu,v∈R2 by the formula (2) and then use the formula (1) to calculate the angle betweenuandv.
Example 4.2.7.Suppose thatu= (√3,1) and v= (√3,3). Then by the formula (2), we have
u·v= 3 + 3 = 6.
Note now that
kuk= 2 and kvk= 2√3.
It follows from the formula (1) that
cosθ= u·v
kukkvk =
6 4√3 =
√
3 2 ,
so thatθ=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 thatuandvare orthogonal.
PROPOSITION 4D.(SCALAR PRODUCT)Suppose thatu,v,w∈R2 andc∈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 ifu=0.
Proof.Writeu= (u1, u2),v= (v1, v2) and w= (w1, w2), whereu1, 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) = (u1v1+u2v2) + (u1w1+u2w2) =u·v+u·w.
Part (c) is rather simple. To check parts (d) and (e), note thatu·u=u2
1+u22 ≥0, and that equality
Consider the diagram below:
P
R A
Q
O
a v
w u
(3)
Here we represent the two vectorsaanduby−→OAand−−→OP respectively. If we project the vectoruon to the lineOA, then the image of the projection is the vector w, represented by−−→OQ. On the other hand, if we project the vectoruon to a line perpendicular to the lineOA, then the image of the projection is the vectorv, represented by−−→OR.
Definition.In the notation of the diagram (3), the vectorwis called the orthogonal projection of the vectoruon the vector a, and denoted by w= projau.
PROPOSITION 4E.(ORTHOGONAL PROJECTION)Suppose thatu,a∈R2. Then
projau= u·a
kak2a.
Remark.Note that the component ofuorthogonal toa, represented by −−→ORin the diagram (3), is
u−projau=u− u·a kak2a.
Proof of Proposition 4E.Note thatw=kafor some k∈R. It clearly suffices to prove that
k= u·a
kak2.
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−ka)·a= 0. Hence
k=u·a
a·a = u·a kak2
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 lineax+by+c= 0. Consider the diagram below:
P
n= (a, b)
Q
O
D
u
Suppose that (x1, y1) is any arbitrary pointOon the lineax+by+c= 0. For any other point (x, y) on the lineax+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 −−→OQrepresents the orthogonal projection projnuofu
on the vectorn. Clearly the perpendicular distanceD of the point (x0, y0) from the lineax+by+c= 0 satisfies
D=kprojnuk=
u·n knk2n
=|(x0−x1, y0√ −y1)·(a, b)| a2+b2 =
|ax0+by0−ax1−by1| √
a2+b2 =
|ax0+by0+c| √
a2+b2 .
We have proved the following result.
PROPOSITION 4F.The perpendicular distanceD of a point (x0, y0)from a line ax+by+c= 0is given by
D=|ax0√+by0+c| a2+b2 .
Example 4.2.9.The perpendicular distanceDof the point (5,7) from the line 2x−3y+ 5 = 0 is given by
D=|10√−21 + 5|
4 + 9 =
6
√
13.
4.3. Vectors inR3
In this section, we consider the same problems as in Section 4.2, but in 3-space R3. Any reader who feels confident may skip this section.
A vector on the planeR3can be described as an ordered tripleu= (u1, u2, u3), whereu1, 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, ifu1=v1,u2=v2 andu3=v3.
Definition.For any two vectorsu= (u1, u2, u3) andv= (v1, v2, v3) inR3, we define their sum to be
u+v= (u1, u2, u3) + (v1, v2, v3) = (u1+v1, u2+v2, u3+v3).
Definition.For any vectoru= (u1, u2, u3) inR3 and any scalar c∈R, we define the scalar multiple to be
cu=c(u1, u2, u3) = (cu1, cu2, cu3).
PROPOSITION 4A’.(VECTOR ADDITION) (a) For everyu,v∈R3, we haveu+v∈R3.
(b) For everyu,v,w∈R3, we haveu+ (v+w) = (u+v) +w.
(c) For everyu∈R3, we haveu+0=u, where0= (0,0,0)∈R3.
(d) For everyu∈R3, there existsv∈R3 such that u+v=0.
(e) For everyu,v∈R3, we haveu+v=v+u.
PROPOSITION 4B’.(SCALAR MULTIPLICATION)
(a) For everyc∈Randu∈R3, we havecu∈R3.
(b) For everyc∈Randu,v∈R3, we have c(u+v) =cu+cv.
(c) For everya, b∈Randu∈R3, we have(a+b)u=au+bu.
(d) For everya, b∈Randu∈R3, we have(ab)u=a(bu).
(e) For everyu∈R3, we have1u=u.
Definition.For any vectoru= (u1, u2, u3) inR3, we define the norm ofuto be the non-negative real number
kuk=
q u2
1+u22+u23.
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 distanced(P, Q) is then the norm of this vector, so that
d(P, Q) =p(v1−u1)2+ (v2−u2)2+ (v3−u3)2.
(2) It is not difficult to see that for any vectoru∈R3 and any scalarc∈R, we have
kcuk=|c|kuk.
Definition.Any vectoru∈R3 satisfyingkuk= 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 inR3.
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 toR3 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 productu·v ofuandv by
u·v=nkukkvkcosθ ifu6=0andv6=0,
0 ifu=0orv=0. (4)
Alternatively, we write
u·v=u1v1+u2v2+u3v3. (5)
PROPOSITION 4C’.Suppose that u= (u1, u2, u3) andv = (v1, v2, v3)are non-zero vectors in R3,
and that θ∈[0, π]represents the angle between them. Then
kukkvkcosθ=u1v1+u2v2+u3v3.
Remarks.(1) We say that two non-zero vectors inR3are orthogonal if the angle between them isπ/2. It follows immediately from the definition of the scalar productthat two non-zero vectorsu,v∈R3 are orthogonal if and only ifu·v= 0.
(2) We can calculate the scalar product of any two non-zero vectorsu,v∈R3 by the formula (5) and then use the formula (4) to calculate the angle betweenuandv.
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 thatkuk= 2 andkvk= 2. It follows from the formula (4) that
cosθ= u·v
kukkvk =
2 4 =
1 2,
so thatθ=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 thatuandvare orthogonal.
The following result is the analogue of Proposition 4D. The proof is similar.
PROPOSITION 4D’.(SCALAR PRODUCT)Suppose thatu,v,w∈R3 andc∈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 ifu=0.
Suppose now thataanduare two vectors inR3. Then since two vectors are always coplanar, we can draw the following diagram which represents the plane they lie on:
P
R A
Q
O
a v
w u
(6)
Note that this diagram is essentially the same as the diagram (3), the only difference being that while the diagram (3) shows the whole ofR2, the diagram (6) only shows part ofR3. As before, we represent the two vectorsaanduby−→OAand−−→OP respectively. If we project the vectoruon to the lineOA, then the image of the projection is the vector w, represented by−−→OQ. On the other hand, if we project the vector uon to a line perpendicular to the line OA, then the image of the projection is the vector v, represented by−−→OR.
The following result is the analogue of Proposition 4E. The proof is similar.
PROPOSITION 4E’.(ORTHOGONAL PROJECTION)Suppose thatu,a∈R3. Then
projau= u·a
kak2a.
Remark.Note that the component ofuorthogonal toa, represented by −−→ORin the diagram (6), is
u−projau=u− u·a kak2a.
4.4. Vector Products
In this section, we shall discuss a product of vectors unique toR3. 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 thex-direction towards they-direction, while the thumb points towards thez-direction. Alternatively, if we imagine Columbus had never lived and that the earth were flat, then taking the x-direction as east and the y-direction as north, then the
z-direction is upwards!
We shall frequently use the three vectorsi= (1,0,0),j= (0,1,0) andk= (0,0,1) inR3.
Definition.Suppose thatu= (u1, u2, u3) andv= (v1, v2, v3) are two vectors in R3. Then the vector productu×vis defined by the determinant
u×v= det
i j k
u1 u2 u3 v1 v2 v3
.
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
u2 u3 v2 v3
i−det
u1 u3 v1 v3
j+ det
u1 u2 v1 v2
k
=
det
u2 u3 v2 v3
,−det
u1 u3 v1 v3
,det
u1 u2 v1 v2
= (u2v3−u3v2, u3v1−u1v3, u1v2−u2v1).
PROPOSITION 4G.Suppose that u= (u1, u2, u3)andv= (v1, v2, v3) are two vectors inR3. 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 u2 u3 v2 v3 ,−det
u1 u3 v1 v3 ,det u1 u2 v1 v2
=u1det
u2 u3 v2 v3
−u2det
u1 u3 v1 v3
+u3det
u1 u2 v1 v2 = det
u1 u2 u3 u1 u2 u3 v1 v2 v3
,
in view of cofactor expansion by row 1. On the other hand, clearly
det
u1 u2 u3 u1 u2 u3 v1 v2 v3
= 0.
This proves part (a). The proof of part (b) is similar.
Example 4.4.1.Suppose thatu= (1,−1,2) andv= (3,0,2). Then
u×v= det
i j k
1 −1 2
3 0 2
=
det
−1 2
0 2
,−det
1 2 3 2 ,det
1 −1
3 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 thatu,v,w∈R3 andc∈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.Writeu= (u1, u2, u3),v= (v1, v2, v3) andw= (w1, w2, w3). To check part (a), note that
det
i j k
u1 u2 u3 v1 v2 v3
=−det
i j k
v1 v2 v3 u1 u2 u3
.
To check part (b), note that
det
i j k
u1 u2 u3
v1+w1 v2+w2 v3+w3 = det
i j k
u1 u2 u3 v1 v2 v3
+ det
i j k
u1 u2 u3 w1 w2 w3
.
Part (c) is similar. To check part (d), note that
cdet
i j k
u1 u2 u3 v1 v2 v3
= det
i j k
cu1 cu2 cu3 v1 v2 v3
= det
i j k
u1 u2 u3 cv1 cv2 cv3
.
To check parts (e) and (f), note that
u×0= det
i j k
u1 u2 u3
0 0 0
= 0 and u×u= det
i j k
u1 u2 u3 u1 u2 u3
= 0
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 non-zero vectors in R3,
and that θ∈[0, π]represents the angle between them. Then
(a) ku×vk2=kuk2kvk2−(u·v)2; and
(b) ku×vk=kukkvksinθ.
Proof.Note that
ku×vk2= (u2v3−u3v2)2+ (u3v1−u1v3)2+ (u1v2−u2v1)2 (7)
and
kuk2kvk2−(u·v)2= (u21+u22+u23)(v12+v22+v32)−(u1v1+u2v2+u3v3)2. (8)
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=kukkvkcosθ.
Combining with part (a), we obtain
ku×vk2=kuk2kvk2− kuk2kvk2cos2θ=kuk2kvk2sin2θ.
Part (b) follows.
Consider now a parallelogram with verticesO, A, B, C. Suppose thatuandvare represented by−→OA
and −−→OC respectively. If we imagine the side OA to represent the base of the parallelogram, so that the base has lengthkuk, then the height of the the parallelogram is given by kvksinθ, as shown in the diagram below:
C B
O A
!v!sinθ v
u
θ
!u!
It follows from Proposition 4J that the area of the parallelogram is given byku×vk. We have proved the following result.
PROPOSITION 4K.Suppose thatu,v∈R3. Then the parallelogram withuandvas two of its sides has areaku×vk.
the magnitude ofu×vdepends only on the norm ofuandvand the angle between the two vectors. It 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 thatu,v,w∈R3do not all lie on the same plane. Consider the parallelepiped withu,v,w as three of its edges. We are interested in calculating the volume of this parallelepiped. Suppose thatu,v
andw are represented by−→OA,−−→OB and−−→OCrespectively. Consider the diagram below:
P A
C
O B
v×w
u
w
v
By Proposition 4K, the base of this parallelepiped, withO, B, Cas three of the vertices, has areakv×wk. Next, note that ifOP is perpendicular to the base of the parallelepiped, then−−→OP is in the direction of
v×w. IfP Ais perpendicular toOP, then the height of the parallelepiped is equal to the norm of the orthogonal projection ofuonv×w. In other words, the parallelepiped has height
kprojv×wuk=
u·(v×w)
kv×wk2 (v×w) =
|u·(v×w)| kv×wk .
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 withu,v andw 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
andw.
Remarks.(1) It follows immediately from Proposition 4L that three vectors inR3 are coplanar if and only if their scalar triple product is zero.
(2) Note that
u·(v×w) = (u1, u2, u3)· det v2 v3 w2 w3 ,−det
v1 v3 w1 w3 ,det v1 v2 w1 w2
=u1det
v2 v3 w2 w3
−u2det
v1 v3 w1 w3
+u3det
v1 v2 w1 w2 = det
u1 u2 u3 v1 v2 v3 w1 w2 w3
, (9)
(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 thatu= (1,0,1), v= (2,1,3) andw= (0,1,1). Then
u·(v×w) = det
1 0 1 2 1 3 0 1 1
= 0,
so thatu,v andware coplanar.
Example 4.5.2. The volume of the parallelepiped withu= (1,0,1), v= (2,1,4) and w= (0,1,1) as three of its edges is given by
|u·(v×w)|=
det
1 0 1 2 1 4 0 1 1
=| −1|= 1.
4.6. Application to Geometry in R3
In this section, we shall study lines and planes inR3 by using our results on vectors inR3.
Consider first of all a plane in R3. Suppose that (x1, y1, z1) ∈ R3 is a given point on this plane. Suppose further thatn= (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)
joins one point on the plane to another point on the plane, and so must be parallel to the plane and hence perpendicular ton= (a, b, c). It follows that the scalar product
(a, b, c)·(x−x1, y−y1, z−z1) = 0,
and so
a(x−x1) +b(y−y1) +c(z−z1) = 0. (10)
If we write−d=ax1+by1+cz1, then (10) can be rewritten in the form
ax+by+cz+d= 0. (11)
Equation (10) is usually called the point-normal form of the equation of a plane, while equation (11) is usually known as the general form of the equation of a plane.
Example 4.6.2.Consider the plane through the points (1,1,1), (2,2,0) and (4,−6,2). Then the vectors
(2,2,0)−(1,1,1) = (1,1,−1) and (4,−6,2)−(1,1,1) = (3,−7,1)
join the point (1,1,1) to the points (2,2,0) and (4,−6,2) respectively and are therefore parallel to the plane. It follows that the vector product
(1,1,−1)×(3,−7,1) = (−6,−4,−10)
is perpendicular to the plane. The equation of the plane is then given by−6(x−1)−4(y−1)−10(z−1) = 0, or 3x+ 2y+ 5z−10 = 0.
Consider next a line inR3. Suppose that (x1, y1, z1)∈R3is a given point on this line. Suppose further thatn= (a, b, c) is a vector parallel to this line. Then for any arbitrary point (x, y, z)∈R3on this line, the vector
(x, y, z)−(x1, y1, z1) = (x−x1, y−y1, z−z1)
joins one point on the line to another point on the line, and so must be parallel ton= (a, b, c). It follows that there is some numberλ∈Rsuch 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 non-zero. Then, eliminating the parameterλ, we obtain
x−x1 a =
y−y1 b =
z−z1
c . (13)
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 4.6.3.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
3 =
y+ 5
5 =−
Example 4.6.4.Consider the line through the points (3,0,5) and (7,0,8). Then a vector in the direction of the line is given by
(7,0,8)−(3,0,5) = (4,0,3).
The equation of the line is then given in parametric form by
x= 3 + 4λ, y= 0, z= 5 + 3λ,
and in symmetric form by
x−3
4 =
z−5
3 and y= 0.
Consider the plane through three fixed points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), not lying on the same line. Let (x, y, z) be a point on the plane. Then the vectors
(x, y, z)−(x1, y1, z1) = (x−x1, y−y1, z−z1),
(x, y, z)−(x2, y2, z2) = (x−x2, y−y2, z−z2),
(x, y, z)−(x3, y3, z3) = (x−x3, y−y3, z−z3),
each joining one point on the plane to another point on the plane, are all parallel to the plane. Using the vector product, we see that the vector
(x−x2, y−y2, z−z2)×(x−x3, y−y3, z−z3)
is perpendicular to the plane, and so perpendicular to the vector (x−x1, y−y1, z−z1). It follows that the scalar triple product
(x−x1, y−y1, z−z1)·((x−x2, y−y2, z−z2)×(x−x3, y−y3, z−z3)) = 0;
in other words,
det
x−x1 y−y1 z−z1 x−x2 y−y2 z−z2 x−x3 y−y3 z−z3
= 0.
This is another technique to find the equation of a plane through three fixed points.
Example 4.6.5. We return to the plane in Example 4.6.2, through the three points (1,1,1), (2,2,0) and (4,−6,2). The equation is given by
det
x−1 y−1 z−1
x−2 y−2 z−0
x−4 y+ 6 z−2
= 0.
The determinant on the left hand side is equal to−6x−4y−10z+ 20. Hence the equation of the plane is given by−6x−4y−10z+ 20 = 0, or 3x+ 2y+ 5z−10 = 0.
We observe that the calculation for the determinant above is not very pleasant. However, the technique can be improved in the following way by making less reference to the unknown point (x, y, z). Note that the vectors
(x, y, z)−(x1, y1, z1) = (x−x1, y−y1, z−z1),
(x2, y2, z2)−(x1, y1, z1) = (x2−x1, y2−y1, z2−z1),
each joining one point on the plane to another point on the plane, are all parallel to the plane. Using the vector product, we see that the vector
(x2−x1, y2−y1, z2−z1)×(x3−x1, y3−y1, z3−z1)
is perpendicular to the plane, and so perpendicular to the vector (x−x1, y−y1, z−z1). It follows that the scalar triple product
(x−x1, y−y1, z−z1)·((x2−x1, y2−y1, z2−z1)×(x3−x1, y3−y1, z3−z1)) = 0;
in other words,
det
x−x1 y−y1 z−z1 x2−x1 y2−y1 z2−z1 x3−x1 y3−y1 z3−z1
= 0.
Example 4.6.6.We return to the plane in Examples 4.6.2 and 4.6.5, through the three points (1,1,1), (2,2,0) and (4,−6,2). The equation is given by
det
x−1 y−1 z−1 2−1 2−1 0−1 4−1 −6−1 2−1
= 0.
The determinant on the left hand side is equal to
det
x−1 y−1 z−1
1 1 −1
3 −7 1
=−6(x−1)−4(y−1)−10(z−1) =−6x−4y−10z+ 20.
Hence the equation of the plane is given by−6x−4y−10z+ 20 = 0, or 3x+ 2y+ 5z−10 = 0.
We next consider the problem of dividing a line segment in a given ratio. Suppose thatx1andx2are
two given points inR3.
We wish to divide the line segment joiningx1andx2 internally in the ratioα1:α2, whereα1 andα2
are positive real numbers. In other words, we wish to find the pointx on the line segment joiningx1
andx2 such that
kx−x1k kx−x2k
=α1
α2,
as shown in the diagram below:
| | |
x1 x x2
←−−−−−− $x−x1$ −−−−−−→←−− $x−x2$ −−→
Sincex−x1 andx2−xare both in the same direction asx2−x1, we must have
α2(x−x1) =α1(x2−x), or x=
α1x2+α2x1 α1+α2 .
We wish next to find the pointxon the line joiningx1andx2, but not betweenx1 andx2, such that
kx−x1k kx−x2k =
whereα1 andα2 are positive real numbers, as shown in the diagrams below for the casesα1< α2 and
α1> α2respectively:
| | | | | |
x x1 x2 x1 x2 x
Sincex−x1 andx−x2are in the same direction as each other, we must have
α2(x−x1) =α1(x−x2), or x=
α1x2−α2x1 α1−α2 .
Example 4.6.7.Letx1= (1,2,3) andx2= (7,11,6). The point
x= 2x2+x1 2 + 1 =
2(7,11,6) + (1,2,3)
3 = (5,8,5)
divides the line segment joining (1,2,3) and (7,11,6) internally in the ratio 2 : 1, whereas the point
x=4x2−2x1
4−2 =
4(7,11,6)−2(1,2,3)
2 = (13,20,9)
satisfies
kx−x1k kx−x2k =
4 2.
Finally we turn our attention to the question of finding the distance of a plane from a given point. We shall prove the following analogue of Proposition 4F.
PROPOSITION 4F’. The perpendicular distance D of a plane ax+by+cz +d = 0 from a point (x0, y0, z0)is given by
D=|ax0√+by0+cz0+d| a2+b2+c2 .
Proof.Consider the following diagram:
P
n= (a, b, c)
Q
O
D
ax+by+cz=0
Suppose that (x1, x2, x3) is any arbitrary pointOon the planeax+by+cz+d= 0. For any other point (x, y, z) on the planeax+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 vectorn= (a, b, c), in the direction −−→OQ, is perpendicular to the planeax+by+cz+d= 0. Suppose next that the point (x0, y0, z0) is represented by the pointP in the diagram. Then the vector
u= (x0−x1, y0−y1, z0−z1) is represented by−−→OP, and−−→OQrepresents the orthogonal projection projnu
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
D=kprojnuk=
u·n knk2n
= |(x0−x1, y0√−y1, z0−z1)·(a, b, c)| a2+b2+c2
=|ax0+by0√+cz0−ax1−by1−cz1| a2+b2+c2 =
|ax0√+by0+cz0+d| a2+b2+c2
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 planeax+by+cz+d= 0 from the origin is
|d| √
a2+b2+c2.
Example 4.6.8.Consider the plane 3x+ 5y−4z+ 37 = 0. The distance of the point (1,2,3) from the plane is
|3 + 10√ −12 + 37|
9 + 25 + 16 = 38
√
50 = 19√2
5 .
The distance of the origin from the plane is
|37| √
9 + 25 + 16 = 37
√
50.
Example 4.6.9.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 4.6.8 that the distance between the two planes is 19√2/5.
4.7. Application to Mechanics
Letu= (ux, uy) denote a vector inR2, where the componentsuxanduy are functions of an independent
variablet. Then the derivative of uwith respect totis given by
du
dt = du
x
dt ,
duy
dt
Example 4.7.1.When discussing planar particle motion, we often letr= (x, y) denote the position of a particle at timet. Then the componentsxandyare functions of t. The derivative
v= dr dt =
dx
dt,
dy
dt
represents the velocity of the particle, and its derivative
a= dv dt =
d2x
dt2,
d2y
dt2
represents the acceleration of the particle. We often writer=krk,v=kvk anda=kak.
Suppose thatw= (wx, wy) is another vector inR2. Then it is not difficult to see that
d
dt(u·w) =u·
dw
dt +
du
dt ·w. (14)
Example 4.7.2. Consider a particle moving at constant speed along a circular path centred at the origin. Thenr =krk is constant. More precisely, the position vectorr = (x, y) satisfies x2+y2 =c1,
wherec1is a positive constant, so that
r·r= (x, y)·(x, y) =c1. (15)
On the other hand,v=kvkis constant. More precisely, the velocity vector
v=
dx
dt,
dy dt satisfies dx dt 2 + dy dt 2 =c2,
wherec2is a positive constant, so that
v·v=
dx
dt,
dy dt · dx
dt,
dy
dt
=c2. (16)
Differentiating (15) and (16) with respect tot, and using the identity (14), we obtain respectively
r·v= 0 and v·a= 0. (17)
Using the properties of the scalar product, we see that the equations in (17) show that the vector v
is perpendicular to both vectors r and a, and so a must be in the same direction as or the opposite direction tor. Next, differentiating the first equation in (17), we obtain
r·a+v·v= 0, or r·a=−v2<0.
Letθdenote the angle betweenaandr. Thenθ= 0◦
orθ= 180◦
. Since
r·a=krkkakcosθ,
it follows that cosθ <0, and so θ= 180◦
. We also obtain ra=v2, so thata=v2/r. This is a vector
proof that for circular motion at constant speed, the acceleration is towards the centre of the circle and of magnitudev2/r.
Letu= (ux.uy, uz) denote a vector in R3, where the components ux, uy anduz are functions of an
independent variablet. Then the derivative ofuwith respect tot is given by
du
dt = du
x
dt ,
duy
dt ,
duz
dt
Suppose thatw= (wx, wy, wz) is another vector inR3. Then it is not difficult to see that
d
dt(u·w) =u·
dw
dt +
du
dt ·w. (18)
Example 4.7.3. When discussing particle motion in 3-dimensional space, we often let r = (x, y, z) denote the position of a particle at time t. Then the components x, y and z are functions of t. The derivative
v= dr dt =
dx
dt,
dy
dt,
dz
dt
= ( ˙x,y,˙ z˙)
represents the velocity of the particle, and its derivative
a= dv dt =
d2x
dt2,
d2y
dt2,
d2z
dt2
= (¨x,y,¨ z¨)
represents the acceleration of the particle.
Example 4.7.4.For a particle of massm, the kinetic energy is given by
T = 12m( ˙x2+ ˙y2+ ˙z2) =12m( ˙x,y,˙ z˙)·( ˙x,y,˙ z˙) = 12mv·v.
Using the identity (18), we have
dT
dt =ma·v=F·v,
whereF=ma denotes the force. On the other hand, suppose that the potential energy is given byV. Using knowledge on functions of several real variables, we can show that
dV
dt = ∂V
∂x
dx
dt + ∂V
∂y
dy
dt + ∂V
∂z
dz
dt = ∂V ∂x, ∂V ∂y, ∂V ∂z
·v=∇V ·v,
where
∇V =
∂V ∂x, ∂V ∂y, ∂V ∂z
is called the gradient ofV. The law of conservation of energy says thatT+V is constant, so that
dT
dt +
dV
dt = (F+∇V)·v= 0
holds for all vectorsv, so thatF(r) =−∇V(r) for all vectorsr.
Example 4.7.5. If a force acts on a moving particle, then the work done is defined as the product of the distance moved and the magnitude of the force in the direction of motion. Suppose that a forceF
acts on a particle with displacementr. Then the component of the force in the direction of the motion is given byF·u, where
u= r
krk
is a unit vector in the direction of the vectorr. It follows that the work done is given by
krk
F· r krk
For instance, we see that the work done in moving a particle along a vectorr= (3,−2,4) with applied forceF= (2,−1,1) isF·r= (2,−1,1)·(3,−2,4) = 12.
Example 4.7.6.We can also resolve a force into components. Consider a weight of mass m hanging from the ceiling on a rope as shown in the picture below:
m
Here the rope makes an angle of 60◦
with the vertical. We wish to find the tensionT on the rope. To find this, note that the tension on the rope is a force, and we have the following picture of forces:
T1 60◦ T2
magnitudemg
The forceT1has magnitudekT1k=T. Letzbe a unit vector pointing vertically upwards. Using scalar
products, we see that the component of the forceT1in the vertical direction is
T1·z=kT1kkzkcos 60◦=12T.
Similarly, the forceT2 has magnitudekT2k=T, and the component of it in the vertical direction is
T2·z=kT2kkzkcos 60
◦
=1 2T.
Since the weight is stationary, he total force upwards on it is 1 2T+
1
Problems for Chapter 4
1. For each of the following pairs of vectors inR2, calculate u+ 3v, u·v,ku−vkand find the angle betweenuandv:
a) u= (1,1) and v= (−5,0) b) u= (1,2) andv= (2,1)
2. For each of the following pairs of vectors inR2, calculate 2u−5v, ku−2vk, u·v and the angle betweenuandv(to the nearest degree):
a) u= (1,3) and v= (−2,1) b) u= (2,0) andv= (−1,2)
3. For the two vectorsu= (2,3) and v = (5,1) in the 2-dimensional euclidean space R2, determine each of the following:
a) u−v b) kuk
c) u·(u−v) d) the angle betweenuandu−v
4. For each of the following pairs of vectors in R3, calculate u+ 3v, u·v, ku−vk, find the angle betweenuandv, and find a unit vector perpendicular to both uandv:
a) u= (1,1,1) and v= (−5,0,5) b) u= (1,2,3) andv= (3,2,1)
5. Find vectorsv and w such thatv is parallel to (1,2,3), v+w= (7,3,5) and w is orthogonal to (1,2,3).
6. LetABCD be a quadrilateral. Show that the quadrilateral obtained by joining the midpoints of adjacent sides ofABCDis a parallelogram.
[Hint: Leta,b,canddbe vectors representing the four sides ofABCD.]
7. Suppose thatu,vandw are vectors inR3 such that the scalar triple porductu·(v×w)6= 0. Let
u′
= v×w
u·(v×w), v
′
= w×u
u·(v×w), w
′
= u×v
u·(v×w).
a) Show thatu′
·u= 1. b) Show that u′
·v=u′
·w= 0.
c) Use the properties of the scalar triple product to findv′
·v andw′
·w, as well as v′
·u,v′
·w,
w′
·uandw′
·v.
8. Suppose thatu,v,w, u′
,v′
andw′
are vectors inR3 such that u′
·u=v′
·v=w′
·w= 1 and
u′
·v=u′
·w=v′
·u=v′
·w=w′
·u=w′
·v= 0. Show that ifu·(v×w)6= 0, then
u′
= v×w
u·(v×w), v
′
= w×u
u·(v×w), w
′
= u×v
u·(v×w).
9. Suppose thatu, vandw are vectors inR3. a) Show thatu×(v×w) = (u·w)v−(u·v)w. b) Deduce that (u×v)×w= (u·w)v−(v·w)u.
10. Consider the three pointsP(2,3,1),Q(4,2,5) andR(1,6,−3). a) Find the equation of the line throughP andQ.
b) Find the equation of the plane perpendicular to the line in part (a) and passing throughR. c) Find the distance betweenRand 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).
11. Consider the points (1,2,3), (0,2,4) and (2,1,3) inR3.
a) Find the area of a parallelogram with these points as three of its vertices.
b) Find the perpendicular distance between (1,2,3) and the line passing through (0,2,4) and (2,1,3).
12. Consider the points (1,2,3), (0,2,4) and (2,1,3) inR3.
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).
13. Find the equation of the plane through the points (1,2,−3), (2,−3,4) and (−3,1,2).
14. Find the equation of the plane through the points (2,−1,1), (3,2,−1) and (−1,3,2).
15. 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.
16. Consider a weight of mass m hanging from the ceiling supported by two ropes as shown in the picture below:
m
Here the rope on the left makes an angle of 45◦
with the vertical, while the rope on the right makes an angle of 60◦
