Vectors and
3-D Geometry
01. Introduction
02. Basic Vector Operations
03. Dot Product
04. Product Product
05. Scalar Triple Product
06. Vector Triple Product
07. More Geometry with Vectors
08. Appendix : 3-D Geometry
CONCEPT NO
CONCEPT NO
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Vectors and
3-D Geometry
In very basic terms, a vector can be thought of as an arrow in the Euclidean plane. This arrow has a starting (initial ) point A and an ending (final) point B:
B
A
Fig - 1
This vector will be represented as !!!"AB. We thus see that a vector has two quantities associated with it:
(a) a magnitude (b) a direction
These two quantities are necessary to carry someone from A to B; these two quantities are sufficient to uniquely specify a vector.
Contrast a vector with a scalar, which is a physical quantity with just a magnitude but no associated direction. Think of a force acting on a block.
F
M
Fig - 2
→
θ
To specify this force, you must specify both its magnitude and direction, and thus force is a vector quantity. You might, as an example, specify this force by saying that it is 10 N strong and is applied at an angle of 30º to the horizontal.
On the other hand, think about the work done by this force over a certain distance, which is obviously a scalar since it will be a quantity with just a magnitude and no direction.
In the discussion that follows, we will see that for a physical quantity to be classified as a vector, it must satisfy another constraint in addition to possessing a magnitude and a direction: it must satisfy the vector law of addition (section – 2). In fact, there do exist quantities (like the rotation of a rigid body) which posses both magnitude and direction but are not vectors because they do not satisfy the addition law.
We can represent a vector using its end-points (like !!!"AB earlier) or we can use lower-case letters (like a"). For any vector a", we have three associated characteristics:
Length : The length (or magnitude) of a" will be denoted by |a"|. Length is obviously a scalar.
Support : This is the line along which the vector a" lies.
Sense : The vector PQ!!!" will have a sense from P to Q along the support of PQ!!!", while that of QP!!!" will be from Q to P along the support of PQ!!!". Thus, the sense of a vector specifies its direction along its support.
Some more terminology is in order before we begin to see the properties of vectors:
(A) Zero vector : A vector of magnitude zero is called a zero vector and is denoted as 0". A zero vector does not really have any direction, since how can you define the direction of a point? We thus assume a zero vector to have any arbitrary direction. In a sense, you may say that the zero vector is not a proper vector. In fact, vectors other than the zero vector are called proper vectors!
(B) Unit vectors: Vector a" is a unit vector if it is of unit length, i.e, if | a" | = 1. If a" is a unit vector, it is
generally denoted as ˆa.
(C) Collinear vectors: These are essentially parallel vectors, i.e, have the same or parallel support.
Some elaboration must be done here: we will encounter, in our study of this chapter, either fixed vectors or free vectors. As the name suggests, a fixed vector has its absolute position fixed with respect to any choosen coordinate system; a free vector is one which can be translated to any position in space, keeping its magnitude and direction fixed.
For example, suppose that O is the origin and A is a fixed point in the coordinate system. Then the vector OA!!!" is fixed because its starting point, O is fixed.
On the other hand, suppose a vector a" corresponds to going 1 unit right and 2 units up in the coordinate system. Then a" is free since it can be translated to anywhere in the
coordinate system; it will still represent going 1 unit right and 2 units up.
When we talk of collinear vectors, it is implied that the vectors being talked about are free vectors. Thus, for two vectors to be collinear, their supports only need to be parallel (and not necessarily the same).
(D) Equal vectors : Two vectors a" and b" are equal if
(i) | | | |a" = b"
(ii) their directions are the same, i.e. their supports are the same OR parallel, and they have the same sense.
It should be evident that when we are saying that two vectors are equal, we implicitly assume that the we talking about free vectors.
(E) Co-initial vectors: Fixed vectors having the same initial point are called co-initial vectors. (F) Co-terminus Fixed vectors having the same ending point are called co-terminus vectors. vectors:
(G) Co-planar vectors: A system of free vectors is coplanar if their supports are parallel to the same plane.
Note that defined this way, two free vectors will always be coplanar. This is because you can always bring these two vectors together to have the same initial point, and then a plane can always be drawn through the two vectors. On the other hand, three free vectors might or might not be coplanar; let us think of this more elaborately. Assume three free vectors a b", " and c". Suppose you bring together a" and b" to have the same
initial point O; you then draw the plane passing through a" and b". Now, when c" is
translated so that its initial point is O, it is not necessary for c" also to lie in the plane that
you drew through a" and b". Thus, a b", and" c" might or might not be coplanar
b O c a Plane through a andb Fig - 3 c a b
does not lie in the plane and b c a O c a b
lies lie in the plane and
(H) Negative of a : The negative of a vector a", denoted by –a", is a vector with the same magnitude as a"
vector but has exactly the opposite direction.
(I) Position vector : The position vector of a point P is a fixed vector which joins the origin of the
reference frame to the point P.
You will be able understand the discussion that follows very clearly only if you try to visualise everything physically. Everything about vectors will then automatically fall in place in your mind.
(A) ADDITION OF VECTORS : TRIANGLE / PARALLELOGRAM LAW
Most of you will already be very familiar with how to add vectors, from your study of physics. Consider two vectors a" and b" which we wish to add. Let
c"= +a b" "
Thus, c" should have the same effect as a" andb" combined. To find the combined effect of a" andb", we place the initial point of b" on the end-point of a" (or vice-versa):
b Fig - 4 → B C A a →
A person who starts at point A and walks first along a" and then along b" will reach the point C. Thus, the
combined effect of a" andb" is to take the person from A to C, i.e, a b"+ =" c" should be the vector AC
!!!" : Fig - 5 b → B C A a → c → →c →a →b = + AC =
In general, we see that to add two vectors, say a"andb", we place the initial point of one of them, say b", at the end-point of the other, i.e., a". The vector a b"+" is then the vector joining the tip of a" to the
end-point of b". This is the triangle law of vector addition. a" and b" can equivalently be added using the parallelogram law; we make the two vectors co-initial and complete the parallelogram with these two vectors as its sides:
Fig - 6 b → A C O →a BC→=→a b = Note that B AC →
The vector OC!!!" then gives us the sum of a" andb".
Fig - 7 b → A C O a → B b → a →+
Note that the triangle and the parallelogram law are entirely equivalent; they are two slightly different forms of the same fundamental principle.
We note the following straightforward facts about addition.
(a) Existence of identity: For any vector a",
0
a"+ =" a"
so that 0" vector is the additive identity.
(b) Existence of inverse: For any vector a",
( )
0a"+ − =a" "
and thus an additive inverse exists for every vector.
(c) Commutativity: Addition is commutative; for any two arbitrary vectors a" andb", a"+ = +b" "b a"
(d) Associativity: Addition is associative; for any three arbitrary vectors a b", and" c",
(
) ( )
a"+ +b" c" = a"+ +b" c"
i.e, the order of addition does not matter.
Verify this explicitly by drawing a vector diagram and using the triangle / parallelogram law of addition.
(B) SUBTRACTION OF VECTORS : An extension of addition
Consider two vectors a" and b"; we wish to find c" such that c"= −a b" "
We can slightly modify this relation and write it as
( )
c"= + −a" b"
and thus subtraction can be treated as addition. To do this, we first reverse the vector b" to obtain −b" and then use the triangle / parallelogram law of addition to add the vector a" and (–b"):
b → a b→ → − a → b → −b b→ a→−
(i) Reverse to obtain b –b
(ii) Add and ( ) to obtain a -b a – b OR
(i) Make and co-initial.a b
(ii) Join the tip of to the tip of to obtain
b a
a – b
Fig - 8
Note that from the triangle law, it follows that for three vectors a b", and" c" representing the sides of a triangle as shown, c → a → Fig - 9 b → we must have 0 a b"+ + =" c" "
In fact, for the vectors a ii, =1, 2... ,n "
representing the sides of an n-sided polygon as shown,
a4 → a3 → a2 → a1 → an → Fig - 10 we must have 1 2 ... n 0 a" + +a" + =a" "
since the net effect of all vectors is to bring us back from where we started, and thus our net displacement is the zero vector.
From any two vectors a" and b", prove that
(i) a"+b" ≤ a" + b" (ii) a b"− ≤" a" + b" (iii) a"+ ≥b" a" − b"
When does the equality hold in these cases?
Solution: Consider this figure: Fig - 11 b → B C A a → – b b→ a→− b → a → + C'
The first two relations follow from the fact that in any triangle, the sum of two sides is greater than the third side:
In ∆∆∆∆∆ABC: AC≤ AB+BC (we’ll soon talk about how and when the equality comes)
⇒ a"+ ≤ +b" a" b"
In ∆∆∆∆∆ABC ': AC'≤AB+BC'=AB+BC
⇒ a"− ≤ +b" a" b"
In the first relation, the equality can hold only if the two vectors have the same direction; this should be intuitively obvious: a → A b → B O Fig - 12 |a + b| = OB→ → = OA + AB = | a | + | b |→ →
The equality in the second relation holds if the two vectors are exactly opposite:
a → A B' O Fig - 13 –b→ b → B A |a – b| = |a + –b→ → ( )| = |OB' | = |OA + AB'| = | a | + | b |→ → → →
To prove the third relation, we use in ∆ABC in Fig - 11, the geometrical fact that the difference of any two sides of a triangle is less than its third side:
AB−BC ≤AC
a b a b
⇒ " − " ≤ +" "
The equality holds when a" andb" are precisely in the opposite direction
a → A O Fig - 14 b → B A |a + b| = |OA + AB|→ → = |OB| = |OA – AB| = || a | – | b ||→ → B
The main point to understand from this example is how easily vector relations follows from corresponding
geometrical facts.
Suppose that the vectors a" andb" represent two adjacent sides of a regular hexagon. Find the vectors representing the other sides.
Solution: Let the hexagon be A1A2A3A4A5A6, as shown:
A3 A2 A1 A6 A5 A4 Fig - 15 a → b →
First of all, we note an important geometrical property of a regular hexagon: Diagonal = 2 × side
⇒ A1A4 = 2 × A2A3 Also, since A1A4 || A2 A3, we have
1 4 A A !!!!!" = 2 × !!!!!"A A2 3 = 2b" Example – 2
Now we use the triangle law to determine the various sides: 3 4 A A !!!!!" = !!!!!" !!!!"A A1 4−A A1 3 = 2b"− +
( )
a" b" = b" "−a 4 5 A A !!!!!"= – a" (only the sense differs; support is parallel to
the support of a") 5 6 A A !!!!!" = −!!!!!"A A2 3 = – b" 6 1 A A !!!!" = −!!!!!"A A3 4 = a b"−"
Thus, all sides are expressible in terms of a" andb".
What can be interpreted about a" andb" if they satisfy the relation:
a"+ = −b" a" b"
Solution: Make a" and b" co-initial so that they form the adjacent sides of a parallelogram:
Fig - 16 b → A C O →a B We have, a"+ =b" OC!!!" =OC and a"− =b" BA!!!" =BA
Thus, the stated relation implies that the two diagonals of the parallelogram OACB are equal, which can only happen if OACB is a rectangle.
This implies that a" and b" form the adjacent sides of a rectangle. In other words, a" andb" are perpendicular to each other.
(C) MULTIPLICATION OF A VECTOR BY A SCALAR
Intuitively, we can expect that if we multiply a vector a" by some scalar λ, the support of the vector will
not change; only its magnitude and / or its sense will. Specifically, if λ is positive, the vector will have the same direction; only its length will get scaled according to the magnitude of λ. If λ is negative, the direction of the product vector will be opposite to that of the original vector; the length of the product vector will depend on the magnitude of λ.
Fig - 17 a → λa→ λ > 0 a → λa→ λ < 0
Note that for any vector a", if we denote the unit vector along a" by ˆa, we have
ˆ
a"= a a"
Put in words, if we multiply the unit vector along a vector a" by its magnitude, we obtain that vector itself.
Put in a slightly different way, we have
ˆ a a a = " "
i.e, if we divide a vector by its magnitude, we obtain the unit vector along that vector’s direction.
Another very important result that follows from this discussion is that two vectors a" and b" are collinear if and only if there exists some λ ∈# such that
a"=λb" Collinear vectors
i.e, two vectors are collinear if one can be obtained from the other simply by multiplying the latter with a scalar.
This fact can be stated in another way : consider two non-collinear vectors a" and b". If for some λ µ, ∈#, the relation
0
a b
λ"+µ" "= ... (1)
is satisfied, then λ and µ must be zero. This is because (1) can be written as
a µ b λ = − " "
which would imply that a" is a scalar multiple of b", i.e., a" and b" are collinear, contradicting our initial supposition that a" andb" are non-collinear.
In subsequent discussions, we’ll be talking a lot about linear combinations of vectors. Let us see what we mean by this. Consider n arbitrary vectors a a" "1, ... .2 a"n A linear combination of these n vectors is a
vector r" such that
1 1 2 2 ... n n
r"=λa" +λa" + +λ a" ... (2)
where λ λ λ ∈1, ....2 n # are arbitrary scalars. Any sort of combination of the form in (2) will be termed a linear combination.
Thus, using the terminology of linear combinations, we can restate the result we obtained earlier: for any two non-zero and non-collinear vectors a" andb", if their linear combination is zero, then both the scalars in the linear combination must be zero.
We now come to a very important concept.
THE BASIS OF A VECTOR SPACE
Consider a two-dimensional plane, and any two arbitrary non-collinear vectors a" and b" in this plane. We make the two vectors co-initial and use their supports as our reference axes:
Fig - 18 b → a →
A reference axes formed by two arbitrary non-collinear vectors
Observe carefully that any vector r" in the plane can be represented in terms of a" andb". We find the components of r" along the directions of a" andb"; those components must be some scalar multiples of a" andb".
Fig - 19 b → a → B C A r → r = OC → = OA + OB = a + bλ→ →µ for some , λ µ #Œ O
Thus, any vector r" in the plane can be written as
r"=λa"+µb" for some λ µ, ∈# ... (1) i.e, any vector r" in the plane can be expressed as a linear combination of a" andb".
We state this fact in mathematical terms as follows: the vectors a" andb" form a basis of our vector space (which is a plane in this case). The term “basis” means that using only a" andb", we can construct any vector lying in the plane of a" andb".
Note that there’s nothing special about a" andb"; any two non-collinear vectors can form a basis for the plane. You must be very clear on the point that two collinear vectors cannot form the basis for a plane while any two non-collinear vectors can. Understanding this fact is very crucial to later discussions.
Try proving this: let a" andb" form the basis of a plane. For any vector r" in the plane of a" andb", we can find scalars λ µ, ∈# such that
r"=λa"+µb" Prove that this representation is unique.
The basic principle that we’ve learnt in this discussion can be expressed in a very useful way as follows:
Three vectors are coplanar if and only if one of them can be expressed as a linear combination of the other two. i.e., three vectors , ,a b c" " " are coplanar if there exist scalars
1, 2
l l ∈ # such that
a"=l b1"+l c2"
We can write this as
(1)a"+ −
( ) ( )
l b1 "+ −l2 c"=0" ⇒ λa"+µb"+γc" =0This form equivalently tells us that three vectors are coplanar if we can find three scalars , ,
λ µ γ ∈# for which their linear combination is zero.
Suppose that for three non-zero vectors a b c", , ," " any two of them are non-collinear. If the vectors
(
a"+2b")
and c" are collinear and the vectors(
b"+3c")
and a" are collinear, prove that2 6 0
a"+ b"+ c"="
Solution: We must have some λ µ, ∈# such that 2 a"+ b"=λc" ... (1) 3 b"+ c"=µa" ... (2) From (1), we have
(
)
1 2 c a b λ = + " " " ... (3) We use this in (2) :(
)
3 2 b a b µa λ + + = " " " " 3 6 1 0 a b µ λ λ ⇒ − + + = " " "Since a" andb" are non-collinear, their linear combination can be zero if and only if the two scalars are zero. This gives 3 0 µ λ − = 6 1 0 λ + = 1 6, 2 λ µ ⇒ = − = −
Using the value of λ in (3), we have
2 6 0
a"+ b"+ c"=
In the preceeding discussion, we talked about the basis of a plane. We can easily extend that discussion to observe that any three non-coplanar vectors can form a basis of three dimensional space:
Fig - 20 r → b → c → R O Q P → r = OS = OP + OQ + OR = a + λ→ µ→ →b + γc for some S , λ µ, γ ∈! a →
In other words, any vector r" in 3-D space can be expressed as a linear combination of three arbitrary non-coplanar vectors. From this, it also follows that for three non-coplanar vectors a b c", , ," " if their linear combination is zero, i.e, if
0
a b c
λ"+µ"+γ" ="
(
where , ,λ µ γ ∈#)
then λ µ, andγ must all be zero. To prove this, assume the contrary. Then, we havea µ b γ c λ λ = − + − " " "
which means that a" can be written as the linear combination of b" and c". However, this would make a b", " andc" coplanar, contradicting our initial supposition. Thus, λ µ, andγ must be zero.
We finally come to what we mean by linearly independent and linearly dependent vectors.
Linearly independent : A set of non-zero vectors a a a1, , ....,2 3 an
" " " "
is said to be linearly independent if
vectors
1 1a 2 2a ... nan 0
λ" +λ " + +λ " ="
implies λ λ1 = 2 =....=λn =0
Thus, a linear combination of linearly independent vectors cannot be zero unless all the scalars used to form the linear combination are zero.
Linearly dependent A set of non-zero vectors a a a" " "1, , ,....,2 3 a"n is said to be linearly dependent if there exist
vectors: scalars λ λ λ1, .... ,2 n not all zero such that,
1 1a 2 2a ... nan 0
For example, based on our previous discussions, we see that
(i) Two non-zero, non-collinear vectors are linearly independent. (ii) Two collinear vectors are linearly dependent
(iii) Three non-zero, non-coplanar vectors are linearly independent. (iv) Three coplanar vectors are linearly dependent
(v) Any four vectors in 3-D space are linearly dependent. You are urged to prove for yourself all these assertions.
Let a b"1, and" c" be non-coplanar vectors. Are the vectors 2a"− +b" 3 ,c a" "+ −b" 2 andc" a"+ −b" 3c" coplanar or non-coplanar?
Solution: Three vectors are coplanar if there exist scalars λ µ, ∈# using which one vector can be expressed
as the linear combination of the other two. Let us try to find such scalars:
(
) (
)
2a"− +b" 3c"=λ a"+ −b" 2c" +µ a"+ −b" 3c"
(
2 λ µ) (
a 1 λ µ) (
b 3 2λ 3µ)
c 0⇒ − − "+ − − − "+ + + "="
Since a b c", ," " are non-coplanar, we must have 2− − =λ µ 0
− − − =1 λ µ 0 3 2+ λ+3µ=0
This system, as can be easily verified , does not have a solution for λ and µ.
Thus, we cannot find scalars for which one vector can be expressed as the linear combination of the other two, implying the three vectors must be non-coplanar.
As an additional exercise, show that for three non-coplanar vectors a b", " and c", the vectors
2 3 , 3 5
a"− b"+ c a" "− b"+ c" and − −2a" 3b"−4c" are coplanar.
RESOLUTION OF A VECTOR IN A GIVEN BASIS
Consider two non-collinear vectors a" andb"; as discussed earlier, these will form a basis of the plane in which they lie. Any vector r" in the plane of a" andb" can be expressed as a linear combination of a" and :b"
Fig - 21 r → b → B O →a A r"=OA!!!" + OB!!!" =λa"+µb" for some λ µ, ∈#
The vectors OA!!!" and OB!!!" are called the components of the vector r" along the basis formed by a" and b". This is also stated by saying that the vector r" when resolved along the basis formed by a" andb", gives the components
and
OA!!!" OB!!!". Also, as discussed earlier, the resolution of any vector along a given basis will be unique.
We can extend this to the three dimensional case: an arbitrary vector can be resolved along the basis formed by any three non-coplanar vectors, giving us three corresponding components. Refer to Fig - 20 for a visual picture.
RECTANGULAR RESOLUTION
Let us select as the basis for a plane, a pair of unit vector iˆ and ˆj perpendicular to each other.
Fig - 22 r → B A j^ O θ i^
Let be the angle thatθ r
→
makes with i^ ⇒ OA =| | cos→r θ
Any vector r" in this basis can be written as
r"=OA OB!!!" !!!"+
=
(
r" cosθ) (
iˆ+ r" sinθ)
ˆj =xiˆ+yjˆwhere x and y are referred to as the x and y components of r".
For 3-D space, we select three unit vectors i jˆ ˆ, and kˆ each perpendicular to the other two.
Fig - 23 r → j^ i^ k ^ O
In this case, any vector r" will have three corresponding components, generally denoted by x, y and z. We thus have
ˆ ˆ ˆ r"= + +xi yj zk
The basis (i jˆ ˆ, ) for the two dimensional case and (i j kˆ ˆ, , ˆ) for the three-dimensional case are referred to as rectangular basis and are extremely convenient to work with. Unless otherwise stated, we’ll always be using a rectangular basis from now on. Also, we’ll always be implicitly assuming that we’re working in three dimensions since that automatically covers the two dimensional case.
MAGNITUDE, DIRECTION COSINES AND DIRECTION RATIOS
Consider a vector
ˆ ˆ ˆ r"= + +xi yj zk as shown in the figure below:
Fig - 24 r y x z Z X Y
The magnitude or r" is simply the length of the diagonal of the cuboid whose sides are x, y and z. Thus
2 2 2
r" = x +y +z ... (1)
Suppose r" makes angles α β, andλ with the X, Y and Z axis, as shown:
Fig - 25 r → α β γ Z X Y
Then the quantities
l = cos α m = cos β n = cos γ
are called the direction cosines of r" (abbreviated as DCs. The DCs uniquely determine the direction of the vector. Note that since
ˆ ˆ ˆ r"= + +xi yj zk
we have cos x= r" α =l r" cos y= r" β =m r" cos z= r" γ =n r"
(
)
2 2 2 2 2 2 2 x y z l m n r ⇒ + + = + + "From (1), this gives
2 2 2 1
l +m +n =
We can also infer from this discussion that the unit vector ˆr along r" can be written as ˆ ˆ ˆ ˆ r xi yj zk r r r + + = = " " " = + +liˆ yjˆ zkˆ
Direction ratios (DRs) of a vector are simply three numbers, say a, b and c, which are proportional to the DCs, i.e
l m n
a = b = c
It follows that DRs are not uniqe (DCs obviously are)
From a set of DRs {a, b, c}, the DCs can easily be deduced:
2 2 2 2 2 2 1 l m n l m n a b c a b c a b c 2 2 2 + + = = = = + + + + 2 2 , 2 2 , 2 2 a b c l m n a b2 c a b2 c a b2 c ⇒ = = = + + + + + + ______________________________________________________________________________________ Before we go on to solving examples involving the concepts we’ve seen till now, you are urged to once again go over the entire earlier discussion we’ve had, so that the “big picture” is clear in your mind.
Show that the vectors iˆ−3jˆ+2 , 2kˆ iˆ−4kˆ−4kˆ and 3iˆ+2ˆj−kˆ are linearly independent.
Solution: Let λ µ γ, , ∈# be scalars such that
(
iˆ− +3ˆj 2kˆ) (
+ 2iˆ−4ˆj−4kˆ) (
+ 3iˆ+2ˆj− =kˆ)
0" λ µ γ(
2 3) (
iˆ 3 4 2) (
ˆj 2 4)
kˆ 0 ⇒ λ+ µ+ γ + − −λ µ+ γ + λ− µ γ− =" 2 3 0 3 4 2 0 2 4 0 + + = ⇒ − − + = − − = λ µ γ λ µ γ λ µ γ ... (1)The determined of the coefficient matrix is
1 2 3
3 4 2 0
2 4 1
− − ≠ − −
Thus, the system of equations in(1) has no solution for λ µ, andγ apart from the trivial solution 0.
= = =
λ µ γ This implies that the three vectors are linearly independent.
Let a b", and" c" be three non-coplanar vectors. Prove that the points A
(
2a"+ −b" c") (
, B 5a"− +b" 2c)
and(
8 3 5)
C a"− b"+ c" are collinear. When we say the point P p
( )
" , we mean the point whose position vector, i.e, thevector drawn from the origin O to that point, OP!!!", is p".
Solution: We have been given the position vectors of three points and we are required to prove that they are
collinear. Let us see what condition must be satisfied in order for three points to be collinear:
A B C O (origin) Fig - 26 Three points , and , whose position vectors are
and respectively, will be collinear if = for some A B C OA, OB OC BC λAB λ ∈ ! → → → → →
Thus, there must be some λ∈# for which BC= AB !!!" !!!" λ
(
)
OC OB OB OA ⇒ !!!" !!!"− =λ !!!" !!!"−(
1)
0 OA OB OC ⇒ λ!!!"− +λ !!!" !!!" "+ =(
2a b c)
(
1)
(
5a b 2c) (
8a 3b 5c)
0 λ λ ⇒ "+ − − +" " "− +" " + "− "+ " ="(
3 3λ) (
a 2λ 2) (
b 3 3λ)
c 0 ⇒ − "+ − "+ − "="Since a b", and" c" are non-coplanar, we have 3 3− λ=0 2λ− =2 0 3 3− λ=0
This consistently gives the solution λ=1, implying A, B and C are collinear.
Let a b", and" c" be three non-coplanar vectors. Prove that the points A
(
2a"+3b"−c")
, B a(
"−2b"−3 ,c")
(
3 4 2)
C a"+ b"− c" and D a
(
"−6b"+6c")
are coplanar.Solution: As in the previous example, we first draw a visual picture to determine when four points can be
coplanar.
Fig - 27 B
A C
D
Draw the vectors and and the plane passing through the two vectors. For to lie in this plane,
AB AD C → → → AC AB AD AC
must be coplanar with and must be expressible as a linear combination
⇒ of AB→and AD→
→ →
Thus, as explained in the figure, we must have some scalars λ µ, ∈# for which
AC= AB+ AD !!!" !!!" !!!" λ µ
(
OC OA) (
OB OA) (
OD OA)
⇒ !!!" !!!"− =λ !!!" !!!"− +µ !!!" !!!"− {O is the origin}(
5 4) (
9 7)
a b c a b c a b c ⇒ "+ − =" " λ − −" "+ " +µ − −" "+ "(
1) (
a 1 5 9) (
b 1 4 7)
c 0 ⇒ + +λ µ "+ + λ+ µ "+ − − λ− µ "="Since a b", and" c" are non-coplanar, we must have 1+ + =λ µ 0
1 5+ λ+9µ =0 1 4+ λ+7µ=0
As can be easily verified, this system has the solution λ = −2,µ =1, implying !!!" !!!"AB AC, and !!!"AD are indeed coplanar.
Thus, the points A, B, C and D are coplanar.
Let A a
( )
" andB b( )
" be two fixed points. Find the position vectors of the points lying on the (extended) line AB which divide the segment internally and externally in the ratio m : n.Solution: We consider internal division; the external division case follows analogously.
Let C c
( )
" be the point which divides AB internally in the ratio m : n.A B O Fig - 28 C m : n → a →c →b
We have, m AC AB m n = + !!!" !!!"
(
)
m(
)
OC OA OB OA m n ⇒ − = − + !!!" !!!" !!!" !!!"( )
m c a b a m n ⇒ − = − + " " " " mb na c m n + ⇒ = + " " "Similarly, the point D d( )" which divides AB externally in the ratio m : n is given by
mb na d m n − = − " " "
A particular case of internal division is the mid-point of A a( )" and B( )b" : the mid-point is , 2
a"+b"
Show by vector methods that the angular bisectors of a triangle are concurrent and find the position of the point of concurrency in terms of the position vectors of the vertices.
Solution: Let the vertices of the triangle by A a
( )
" , ( ) and ( )B b" C c" . We use the geometrical fact that an anglebisector divides the opposite side in the ratio of the sides containing the angle.
Fig - 29 y z I D C c( ) B b( ) A a( ) x
Let the sides be of lengths , and . Draw the bisectors of angles and . Suppose they meet in .
x y z
A B
I Example – 10
We thus have,
BD c
DC =b
Thus, D is given by (the internal division formula):
zc yb D z y + ≡ + " "
In∆ ABD, since BI is the angle bisector, we have z x DI BD z y x IA BA z z y + = = = +
Thus, we now have the position vectors of A and D we know what ratio I divides AD in. I can now be easily determined using the internal division formula:
(
)
zc yb xa x y z y I x z y + + + + ≡ + + " " " xa yb zc x y z + + = + + " " " ... (1)The symmetrical nature of this expression proves that the bisector of C will also pass through I. The angle bisectors will therefore be concurrent at I, called the incentre. The position vector of the incentre
is given by (1).
In a parallelogram ABCD, let M be the mid-point of AB. AC and MD meet in E. Prove that both AC and MD are trisected at E. Solution: C D A M B E
We need to prove that
AE EC = ME ED= 1 2 Fig - 30 Example – 11
There’s no loss of generality in assuming A to be the origin (0)" . Let B and C have the position vectors and
b" c". The position vector of D is given by AD !!!" where AD= AC+CD !!!" !!!" !!!" = −c" !!!"AB = −c b" " D is therefore the point c b−
" " .
Since M is the mid-point of AB, M's position vector is 2
b"
. Let us first find the position vector of a point E' which lies on AC and trisects it. We’ll then show that same point lies on DM and trisects on DM and trisects it too, proving the stated assertion.
1 2 0 ' 3 3 c c E ≡ × + × = " " " A point E'' which trisects MD is
( )
2 1 2 ' 3 3 b c b c E × + × − ≡ = " " " "Since E' and E'' are the same-point, say E, we see that AC and MD are trisected at E.
Prove that the lines joining the vertices of a tetrahedron to the centroids of the opposite faces are concurrent.
Solution: It is in this example that the powerful nature of vector algebra will become apparent; this is a 3-D
problem and any other methods of proving the assertion will be extremely cumbersome. For those of you who are more mathematically inclined, you can try proving the assertion using already known methods.
First of all, we need to know how to write the position vector of the centroid of a triangle in terms of the position vectors of its vertices.
Fig - 31 E F G D C c( ) B b( ) A a( )
Recall that the centroid divides any median in the ration 2 : 1 Thus, AG GD= 1 2 Example – 12
Since D is the mid-point of BC, we have
2
b c D≡ +
" "
Since AG : GD = 2 : 1, we can now determine G:
2 1 2 2 1 3 b c a a b c G + × + × + + ≡ = + " " " " " "
We now consider a tetrahedron, say, with the vertices A a
( )
" , ( ),B b" C c( )
" and D d( )" .Fig - 32 C c( ) A a( ) D d ( ) G BCD is the centroid of ∆ G B b( )
Consider the centroid of ∆BCD, say G, which will be given by
3
b c d G≡ + +
" " "
Now, pause and think about the assertion we are required to prove. If the four lines are indeed concurrent, the point of concurrency must be given by a position vector which has a ‘symmetrical’ expression with respect to all the four vertices. Can you find a point on the segment AG which has a symmetrical expression with respect to the four vertices?
A little thought will show that the answer is yes: consider the point P which divides AG in the ratio 3: 1. P will be given by
3 1 3 3 1 4 b c d a a b c d P + + × + × + + + ≡ = + " " " " " " " "
It is immediately apparent now that P lies on each of the four lines joining the vertices to centroids of the opposite faces! In addition, we’ve also been able to find the position vector of P in terms of the position vectors of the four vertices of the tetrahedron.
In a quadrilateral PQRS, PQ!!!"=a QR",!!!"=b" and SP!!"= −a" b". If M is the mid-point of QR and X is a point on SM such that SX : SM = 4 : 5, prove that P, X and R collinear.
Solution: Since no position vectors have been specified in the question (only the sides have been specified),
there is no loss of generality in assuming that P is the origin 0".
M R a + b( ) Q a( ) P O( ) S b – a( ) X
Note how the position vectors of the vertices of the quadrilateral have been specified. is a point such that : = 4 : 5
(Diagram not to scale) Thus, : = 4 : 1 X SX SM SX XM Fig - 33 We have,
( )
2 2 a a b b M ≡ + + = +a " " " " "( )
4 1 2 3 3 4 1 5 b a b a a b X × + + × − + ⇒ ≡ = + " " " " " "( )
3 5 X a b ⇒ ≡ "+" Thus, 3 5 PX = PR !!!" !!!"implying that P, X and R are collinear.
It is known that in a ∆ABC with centroid G, circumcentre O and orthocentre H,
: 1: 2
OG GH =
Let P be any point in the plane of ∆ABC. Prove the following assertions:
(a) !!!" !!!" !!!" !!!!"OG OB OC+ + =OH (b) !!!" !!!" !!!"HA+HB+HC=2HO!!!" (c) PA!!!" !!!" !!!"+PB+PC =3!!!"PG Solution: Fig - 34 D E O F B A C
Let , and be the mid-points of , and respectively D E F BC CA AB G H (a) OA OB OC!!!" !!!" !!!" !!!"+ + =OA+
(
OB OC!!!" !!!"+)
2 OA OD =!!!"+ !!!" (Since D is BC' s mid-point)= 3OG!!!" (Since G lies on AD and divides it in the ratio 2 : 1)
OH
=!!!!" (Since O, G and H are collinear and OH = 3OG)
(b) !!!" !!!" !!!" !!!"HA HB+ +HC=HA+
(
HB!!!" !!!"+HC)
2
HA HD
=!!!"+ !!!" 3HG
= !!!" (Same logic as above) 2
3
3HO
= × !!!" (again, same as above) 2HO
= !!!"
(c) For any arbitrary point P in the plane of ∆ABC, we have
(
)
PA PB+ +PC=PA+ PB+PC !!!" !!!" !!!" !!!" !!!" !!!" 2 PA PD =!!!"+ !!!" 3PG = !!!"Go over the solution again if you find any part of it confusing.
Justify the following tests for collinearity and coplanarity
(a) Three points with position vectors a b c", ," " are collinear iff there exist scalars p, q, r not all zero such that 0
pa"+qb"+rc"=" where p+ + =q r 0
(b) Four points with position vectors a b c d", , ," " " are coplanar iff there exist scalars p, q, r, s not all zero such that
0 pa"+qb"+ +rc" sd" "= where p+ + + =q r s 0
Solution: There is nothing new in these tests; we’ve already seen their justification in the discussion preceeding
the examples. These tests are just the same facts put into slightly different language.
(a) Since p + q + r = 0, we have r = – ( p + q )
Assume r ≠0 Now, pa"+qb"+rc"=0"
(
)
0 pa qb p q c ⇒ "+ "− + "=" pa qb c p q + ⇒ = + " " "This implies that c" is the position vector of a point which divides the points a" andb" in the ratio p : q. Thus, the points a b", and" c" are collinear.
Now, we prove the other way implication, i.e, we first assume that points a b", and" c" are collinear:
(
)
b a λ c a ⇒ " "− = " "−(
1 λ) ( ) ( )
a 1 b λ c 0 ⇒ − "+ − "+ "=" 0 pa qb rc ⇒ "+ "+ "=" where p+ + = − + − +q r(
1 λ) ( ) ( )
1 λ =0 This completes our proof.(b) Again, first assume the existence of scalars p, q, r, s such that 0 p+ + + =q r s
(
)
s p q r ⇒ = − + + Now, 0 pa"+qb"+ +rc" sd" "=(
)
0 pa qb rc p q r d ⇒ "+ "+ −" + + " "=( ) ( ) ( )
0 p a d q b d r c d ⇒ "− " + " "− + "− " =" Assuming p≠0, we have( )
q( )
r( )
a d b d c d p p − = − − + − − " " " " " "( ) ( )
b d c d λ µ = " "− + "− "That
( )
a"−d" can be written as a linear combination of the vectors( )
b"−d" and( )
c"−d" impliesthat
( )
a"−d" ,( )
b"−d" and( )
c"−d" are coplanar.⇒ Points with position vectors a b c d", , ," " " are coplanar.
Now lets prove the other way implication. Assume that A a
( )
" ,B b( )
" ,C c( )
" andD d( )
" are coplanarpoints. Thus, there must exist scalars λ µ, such that AB=λAC+µAD !!!" !!!" !!!"
( )
b a λ(
c a)
µ( )
d a ⇒ "−" = " "− + "−"(
λ µ 1) ( ) ( ) ( )
a 1 b λ c µ d 0 ⇒ + − "+ "+ − "+ − " "= 0 pa qb rc sd ⇒ "+ "+ +" " "= where p+ + + =q r s(
λ µ+ − +1) ( ) ( ) ( )
1 + − + −λ µ = 0If any point O inside or outside a tetrahedron ABCD is joined to the vertices and AO, BO, CO, DO are produced so as to cut the planes of the opposite faces in P, Q, R, S respectively, prove that
1
OP OQ OR OS AP+ BQ+CR+DS =
Solution: Assume O to be the origin, and the position vectors of A, B, C, D to be a b c d", , ," " " respectively. Since a b c d", , ," " " are non-coplanar vectors, we must have scalars λ λ λ λ1, , ,2 3 4 such that (Page 17)
1a 2b 3c 4d 0
λ"+λ "+λ "+λ " "= ... (1)
Since AO is produced to meet the plane of the opposite face in P, !!!"AO andOP!!!" must be collinear vectors. Thus, AO=µOP !!!" !!!" for some µ∈# a µOP ⇒ − =" !!!"
=µp" (p" is the position vector of P) This when used in (1) gives
(
−µλ1)
p+λ2b+λ3c+λ4d =0 " " " " "However, since B, C, D and P will be coplanar, we have, using the result of the last example,
1 2 3 4 0 µλ λ λ λ − + + + = 1 2 3 4 λ µ λ λ λ ⇒ = + + 1 OP OP AP AP µ µ ⇒ = = + !!!" !!!" (From (2)) 1 1 2 3 4 λ λ λ λ λ = + + + 1 i λ λ = ∑ We similarly have, 3 2 , , 4 i i i OQ OR OS BQ CR DS λ λ λ λ λ λ = = = ∑ ∑ ∑ 1 OP OQ OR OS AP BQ CR DS ⇒ + + + = Example – 16
Q. 1 If the points with position vectors 60iˆ+3 , 40ˆj iˆ−8ˆj and aiˆ−52ˆj are collinear, find a.
Q. 2 Let a b c", ," "be three non-coplanar vectors. Prove that the points with position vectors a"−2b"+3 ,c" 2a"+3b"−4c" and − +7b" 10c" are collinear.
Q. 3 Find a point P within a quadrilateral ABCD such that 0
PA PB+ +PC+PD=
!!!" !!!" !!!" !!!" "
Q. 4 ABCD is a parallelogram. Let L and M be the mid-points of BC and CD respectively. Prove that 3
2 AL+AM = AC !!!" !!!!" !!!"
Q. 5 Prove that the line segments joining the mid-points of the adjacent sides of a quadrilateral form a
parallelogram.
Q. 6 Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel
sides and equal to half of their difference.
Q. 7 Let ABCD be a quadrilateral and E and F be the mid-points of AC and BD respectively. Prove that 4
AB+AD+CB+CD= EF !!!" !!!" !!!" !!!" !!!"
Q. 8 Let a b c", ," " be non-coplanar vectors. Prove that the three vectors 3a"−7b"−4 ,c a" "+ +b" 2c" and 3a"−2b"+c" are coplanar
Q. 9 Let a b c", ," " from a linearly independent system of vectors. Show that the system of vectors ma"+ +b" c",
a"+mb"+c" and a b"+ +" mc" form a linearly independent system iff m = –2.
Q. 10 In ∆ABC, the point D lies on AC such that AD : DC = 2 : 1. BD is produced to F such that DF = 2BD . Prove that AF is parallel to BC and is equal to 2BD.
We have already seen the addition and subtraction of vectors. In this section, we’ll understand how we can define the product of two vectors.
Before formally defining the dot product, let us try to understand why (such a) product is required at all. Consider a force F" acting on a block M at an angle θ to the horizontal.
F→
M θ
Fig - 35
This block, as an effect of the force, is displaced through a horizontal distance s. We can denote the displacement by the vector s", which has a horizontal direction and has a magnitude s.
F→ M θ Fig - 36 F→ s →
In physics, you must have studied the concept of the work done by a force. This work done is the maximum if the force and the displacement caused by it are in the same direction, and zero if the force and the displacement are perpendicular. This suggests that we must consider the component of F" along s" to evaluate its work done.
From the figure above, the component of F" along the direction of s" is F" cosθ. The work done will then be cos
W = F s" " θ ...(1)
We denote the right hand side of (1) by the dot product notation. cos
W = F s" " θ = ⋅F s" "
In general, for two vectors a" and b" inclined at an angle θ to each other, their dot product P is defined as
cos θ
P= ⋅ =a b" " a b" "
It is very important to understand that P is a scalar quantity.
You can think of P in this way: it is a measure of the “effect” of one vector along the other. For two vectors of fixed magnitude , their dot product will decrease in magnitude as θ increases from 0 to π
2 (or decreases from to π π 2). We can write P as
(
cos)
(
cos)
P= a b" " θ = a" θ b"Thus, P is the product of the modulus of either vector and the projection of the other in its direction. From the definition of the dot product, we can make certain useful observations about its properties.
(i) The angle θ between two vectors a" and b" is given by
cos a b a b ⋅ θ = " " " "
(ii) a b"⋅ ≤" a b" " , the equality holding only if θ =0 or π (iii) The projection of a" on b" is
ab a b b p a a b b b ⋅ = = ⋅ = ⋅ " " " " " $ " "
(iv) The projection of b" on a" is
$ ba a b a p b a b a a ⋅ = = ⋅ = ⋅ " " " " " " "
(v) Scalar product is commutative, i.e,
a b"⋅ = ⋅" b a" "
(vi) Scalar product is distributive, i.e
( )
a b"⋅ + = ⋅ + ⋅" c" a b" " a c" "
and
( )
a"+ ⋅ = ⋅ + ⋅b" c" a c" " b c" "(vi) The scalar product of two vectors is zero if and only if the two vectors are perpendicular.
This also gives
ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 0
(vii) For any vector a" 2 a a" "⋅ = a" Thus, ˆ ˆ ˆ ˆ ˆ ˆ 1 i i⋅ = ⋅ = ⋅ =j j k k (viii) a"±b"2 = ± ⋅ ±(a" b") (a" b") 2 2 2( ) a b a b = " + " ± "⋅" 2 2 (a b a b"+")("− =") a" −b"
(ix) This property is very important. If two vectors a" and b" have been specified in rectangular form, i.e., 1ˆ 2 ˆ 3 ˆ and 1ˆ 2 ˆ 3 ˆ
a"=a i+a j a k+ b"=b i +b j b k+ then
we have,
(
1ˆ 2ˆ 3ˆ)(
1ˆ 2ˆ 3ˆ)
a b"⋅ =" a i+a j a k b i+ +b j b k+
=a b1 1+a b2 2+a b3 3 {Using properties (vi) and (vii)}
⇒ a b⋅ =a b1 1+a b2 2+a b3 3 "
"
The angle θ between the two vectors will be given by cos a b :
a b ⋅ θ = " " " " 1 1 2 2 3 3 2 2 2 2 2 2 1 2 3 1 2 3 cos a b a b a b a a a b b b + + ⇒ θ = + + + +
(x) The direction cosines l, m, n of a vector a" will be given by ˆ
ˆ ˆ
ˆ , ˆ , ˆ
l = ⋅a i m= ⋅a j n= ⋅a k
(xi) Let r" be a vector coplanar with the vectors a" and b". If r a" "⋅ =0 and r b"⋅ =" 0, this would imply that r" is perpendicular to both a" and b". This can only happen if a" and b" are collinear.
Analogously, let r" be an arbitrary vector and a b c", ," " be three vectors such that 0
r a r b r c" " "⋅ = ⋅ = ⋅ =" " "
This means that r" is perpendicular to each of a", b" and c" which can only happen if a b", " and c" are coplanar.
(xii) Let a b c", ," " be three non-coplanar vectors. We’ve already discussed that a b c", ," " can form a basis for 3-D space. Any vector r" can be written in this basis as
2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) r r a a r b b r c c r a r b r c a b c a b c = ⋅ + ⋅ + ⋅ ⋅ ⋅ ⋅ = + + " " " " " " " " " " " " " " " "
This representation is of significant importance and you must understand how it comes about.
Find the component of a vector b" perpendicular to the vector a".
Solutions: We need to find r", the component of b" perpendicular to a"
R Fig - 37 b → r → Q P a → We have 2 b a PQ a a ⋅ = " " !!!" " " r b PQ ⇒ "= −" !!!" 2 a b b a a ⋅ = − " " " " "
Let a b", " and c" be three mutually perpendicular vectors of equal magnitude. Prove that the vector a b"+ +" c" is
equally inclined to each of the three vectors.
Example – 16
Solutions: Let θa represent the angle between a " and a b"+ +" c". We have, ( ) cos a a a b c a a b c ⋅ + + θ = + + " " " " " " " " 2 a a a b a c a a b c ⋅ + ⋅ + ⋅ = + + " " " " " " " " " " ...(1)
Let the magnitude of a b", " and c" be λ. Also since the three vectors are mutually perpendicular, we
have a b"⋅ = ⋅ =" a c" " 0. Now, 2 ( ) ( ) a"+ +b" c" = + + ⋅ + +a b" " c" a" b" c" = ⋅ + ⋅ + ⋅a a b b" " " " c c" " = λ3 2
Using these facts in (1), we have 2 1 cos 3 3 a λ θ = = λ⋅ λ
It is easy to see that cosθb and cosθc will have the same value. Thus, a b+ +c
" " "
is equally inclined
to a b", " and c".
Find the angle between the two diagonals of a cube.
Solutions: y x z A O B Fig - 38 P
Let us find the angle between the diagonals and . Note that the position vectors of , and are respectively
OP AB
A B P
A ≡ a i^ B ≡ a j + a k^ ^ P ≡ a i + a j + a k^ ^ ^
where is the side of the squarea
We now have, ˆ ˆ ˆ 3 OP!!!"≡ + +ai aj ak⇒ OP!!!" = a AB=OB OA− !!!" !!!" !!!" ≡ − + +aiˆ ajˆ akˆ⇒ !!!"AB = 3a
Let θ denote the angle between OP and AB. Thus,
cos OP AB OP AB ⋅ θ = !!!" !!!" !!!" !!!" ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( 3 ) ( 3 ) ai aj ak ai aj ak a a + + ⋅ − + + = 2 2 2 2 3 a a a a − + + = 1 3 = cos 11 3 − ⇒ θ =
This is the angle between any two diagonals of (any) cube.
Let AD, BE and CF be the medians in ∆ABC. Prove that 0 BC AD CA BE⋅ + ⋅ +AB CF⋅ = !!!" !!!" !!!" !!!" !!!" !!!" Solutions: D C c( ) B b( ) Fig - 39 → → A a( )→ F E Example – 19
Since D is the mid-point of BC, we have
D≡ b2+c
" "
(Position vector of ) (Position vector of )
AD D A ⇒ !!!"= " − " 2 b c a + = − " " " 1( 2 ) 2 b c a = " "+ − " Thus, 1( ) ( 2 ) 2 BC AD⋅ = c− ⋅ + −b b c a !!!" !!!" " " " " " 1
(
2 2 2 ( ))
2 c b a b c = " − " + "⋅ −" " ...(1) Similarly, 1(
2 2 2 ( ))
2 CA BE!!!" !!!"⋅ = a" − c" + b"⋅ −c" "a ...(2) 1(
2 2 2 ( ))
2 AB CF⋅ = b − a + c⋅ −a b !!!" !!!" " " " " " ...(3)It is now immediately apparent that the right hand sides (1), (2) and (3) sum to zero. Thus, the stated
assertion is true.
Prove that the altitudes in a triangle are concurrent.
Solution: Assume the three vertices of the triangle to be A, B and C.
D C c( ) B b( ) Fig - 40 → → A a( )→ E H(0)→
Draw the altitudes and and suppose they intersect in . If we prove that
is perpendicular to , our task will be accomplished. AD BE H CH AB Example – 20
Assume H to be the origin 0" and A, B, C to have the position vectors a b c", ," ". Since AH ⊥BC, we have ( ) 0 a b"⋅ − =" c" a b a c ⇒ "⋅ = ⋅" " " ...(1) Similarly, since BH ⊥AC, ( ) 0 b" " "⋅ −c a = a b b c ⇒ "⋅ = ⋅" " " ...(2)
From (1) and (2), we have
a c" "⋅ = ⋅b c" "
(a b) c 0 ⇒ "− ⋅ =" "
CH AB
⇒ ⊥
Thus, the altitude through C passes through H, implying that the three altitudes are concurrent.
If a, b, c are the lengths of the sides of ∆ABC opposite to the angles A, B and C respectively, prove using vector methods that
(1 cos ) (1 cos ) (1 cos ) ( )(cos cos cos )
a + A +b + B +c + C = + +a b c A+ B+ C Solutions: c → a → Fig - 41 b → B C
Denote for vectors representing the sides of the triangle by
a, b and . c
→ → →
A
We have, by the triangle law,
0 a b"+ + =" c" " ( ) a b c ⇒ "= − +" " ( ) a a a b c ⇒ " "⋅ = − ⋅ +" " " Example – 21
