Fundamental Theorems of Vectors

In this section, we shall discuss some of the fundamental theorems of Vectors in both two and three dimensions. Here, the letters in boldfaced characters denote the vectors i.e. ‘a’ stands for the vector a.
 

Fundamental Theorem of Vectors in Two-Dimensions

If a and b be two non-zero non-collinear vectors, then any vector r in the plane of a and b can be expressed uniquely as a linear combination of a and b i.e. there exist unique l, m ∈ R such that la +mb =r.

This also means that if  l₁a + m₁b = l₂a + m₂b then l₁ = l₂ and m₁ = m₂.
 

Collinear, Coplanar and Parallel Vectors

Collinear Vectors: The word collinear is used to describe vectors which are scalar multiples of one another. These vectors are parallel and can have different magnitudes in the same or opposite directions.  

Coplanar Vectors: The word coplanar is used for vectors which are in at least 3 space. As the name suggests, coplanar vectors are three or more vectors which lie in the same plane which could be any two dimensional flat surface.

In other words, the vectors which lie in the same plane or parallel to the same plane are called coplanar. This usually occurs if either the vectors are parallel or they intersect each other. All collinear vectors will also be coplanar. But all coplanar will not be collinear vectors. This point should be remembered.

Parallel Vectors: When 2 vectors act along the same line but have a separation between them, they are said to be parallel. i.e. Parallel vectors have the same phase, but different magnitudes. Also, it is due to this only that when two vectors a and b are parallel, a= kb, where k is a constant.

--------A---------> Here A and B are parallel.
----------B--------->

Remark: Collinear Vectors are also parallel vectors except that they lie on the same line. When two vectors are parallel, the dot product of the vectors is 1 and their cross product is zero. 
 

Fundamental Theorem of Vectors in Three Dimensions

If a, b and c be three non-zero, non-coplanar vectors in space, then any vector r in space can be expressed uniquely as a linear combination of a, b and c.i.e there exist unique l, m, n ∈ R such that la + mb + nc = r

This also means that if l₁a + m₁b+ n₁c = l₂a+ m₂b+ n₂c, then

l₁ = l₂, m₁ = m₂ and n₁ = n₂.
 

Linear Combination of Vectors

The linear combination of a finite set of vectors a₁,a₂,…a_n is defined as a vector r such that 
r = k₁a₁+ k₂a₂ + ……+ k_n a_n , where k₁, k₂, … k_n are any scalars (real numbers).

Examples

Write the vector m = (1, 2, 3) as a linear combination of the vectors: u = (1, 0, 1), v = (1, 1, 0) and w = (0, 1, 1).

Solution: Let us write the vector m = xu+yv+zw

Then (1,2,3)= x(1,0,1) + y(1,1,0) + z(0,1,1)

(1, 2, 3)= (x+ y, y +z, z +x)  

From these we get three equations

x+y=1

y+z=2

x+z=3

Adding these three we get, 2x+2y+2z=6

So x+y+z=3.

The first set gives z=2.

The second yield x=1.

From the third we get y=0.

Hence the vector m can be written as a linear combination of vectors u, v and w.

To get an idea about the types of questions asked, you may go through the Papers of Previous Years.

Linearly Dependent and Independent  Vectors

 A system of vectors a₁,a₂,…a_n is said to be linearly dependent if there exists a system of scalars k₁, k₂ …, k_n (not all zero) such that k₁a₁+ k₂a₂ + ……+ k_n a_n= 0

They are said to be linearly independent if every relation of the type

k₁a₁+ k₂a₂ + ……+ k_n a_n= 0 implies that k₁ = k₂ =….= k_n = 0.

Notes:

• Two collinear vectors are always linearly dependent.

• Two non-collinear non-zero vectors are always linearly independent.

• Three coplanar vectors are always linearly dependent.

• Three non-coplanar non-zero vectors are always linearly independent.

• More than 3 vectors are always linearly dependent.

• Three points with position vectors a,b,c are collinear if

l₁a+l₂b+l₃c = 0 with l₁ +l₂+l₃ = 0.

• Four points with position vectors a,b,c and d are coplanar if

l₁a+l₂b+l₃c+l₄d with l₁ + l₂ + l₃ + l₄= 0.
 

Watch this Video for more reference
 

Illustration 0 : If a,b and c are non–zero non coplanar vectors, determine whether the vectors r₁= 2a-3b+c, r₂= 3a-5b+2c and r₃= 4a-5b+c are linearly independent or dependent.

Key concept:  Three vectors are linearly dependent if they are coplanar that means any one of them can be represented as a linear combination of other two.

Solution: Let r₃= xr₁+yr₂, where x and y are scalars.

If the given vectors are linearly dependent then x and y will exist uniquely; otherwise not.

Consider r₃= xr₁+yr₂

Then 4a-5b+c = x (2a-3b+c) + y(3a-5b+2c)

= (2x+3y)a + (-3x-5y) b + (x+2y)c

But a, b and c are non-zero and non-coplanar vectors.

Hence, 2x+3y = 4    ….(i)

–3x – 5y = –5           …(ii)

  x + 2y = 1               …(iii)

 Solving (i) and (ii), we get x = 5, y = –2 which also satisfy (iii).

Now x and y are unique so r₃= 5r₁-2r₂.

Hence r₁, r₂ and r₃ are linearly dependent vectors.

Solution: Since a, b and c are given to be linearly dependent vectors, so this also implies that the vectors are coplanar.

So, a. (b x c) = 0

Solving this with the help of determinants, and applying the column operation

C₃ – C₁, we get

(β-1) (3-4) = 0.

This gives β = 1.

Also, |c| = √3

√ 1+ α² + β² = √3

√ 1+ α² +1 = √3

√α² + 2 = √3

α² + 2 = 3 so α = ±1.

Illustration 2: If a, b and c are non-planar unit vectors such that

a x (b x c) = (b + c)/√2, then the angle between a and b is

1. 3π/4                                                               2. π/4

3. π/2                                                                  4. Π

Solution: Since a x (b x c) = (b + c)/√2

Hence, (a.c) b – (a.b) c = b/√2 + c/√2

(a.c) = 1/√2 (since b and c are not coplanar)

(a.b) = - 1/√2

cos θ = - 1/√2 (since a and b are unit vectors)

cos 3π/4 = cos θ

θ = 3π/4

Illustration 3: A₁, A₂, …… , A_n are the vertices of a regular plane polygon with n sides and O is its center. Show that if the summation runs from 1 to (n-1) then

Σ (OA_i x OA_i+1) = (1-n) (OA₂ x OA₁)

Solution: Let v be a unit vector which is perpendicular to the plane of the polygon.

If l is some scalar,

(OA_i x OA_i+1) = lv, where i = 1, 2, 3, …. , (n-1)

L.H.S = (n-1) lv

Thus, R.H.S = (n-1) OA₁ x OA₂ (n-1) lv

Hence, L.H.S = R.H.S 

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