Fundamental Theorem of Vectors

Vectors are defined by magnitude and direction. Vectors can be define as, the line segment AB is a vector which is denoted by a\vec a and is read as vector a. Point A is called the initial point from where AB starts, point B is called the terminal point where AB ends and the distance between A and B is called as magnitude or length of the vector a\vec a and is denoted by a|\vec a|. The arrow indicates the direction of the vector.

Position Vectors:

The position vector of a point P(x,y,z) with respect to origin O (0,0,0) is drawn. Its magnitude will be given by|OP| = x2+y2+z2\sqrt{x^2+y^2+z^2} and the vector OP\vec {OP} will have O and P as its initial and terminal points, respectively. This vector denotes the position vector of the point P with respect to O.

Position Vector Representation

Types of Vectors

  • Zero Vector: A vector which has got the same initial and ending points. It is denoted by O\vec O
  • Unit Vector: A vector with its magnitude equal to one is called a unit vector. Thus, i^\hat i is a unit vector of I -> where i^|\hat i| = 1.
  • Co-initial vectors: When two or more vectors have the same initial points
  • Equal Vectors: Two vectors are equal when they have same magnitude and direction. The initial points doesn’t matter here. They may be represented as: b\vec b = a\vec a
  • Negative of a Vector: A vector whose magnitude is the same but direction is opposite to the original vector say PQ\vec PQ is called a negative vector of O\vec O i.e. QP\vec QP = – PQ\vec PQ
  • Free Vectors: The vectors whose initial points are not fixed.
  • Parallel Vectors: The vectors which may have different magnitude but all should have the same or opposite direction are called parallel vectors.
  • Collinear Vectors: Vectors which may have same direction or are parallel or anti-parallel. As magnitudes can vary, we can find some scalar vector λ for which a=λb\vec a = \lambda \vec b.
  • Non-collinear Vectors: Two vectors acting in different directions are called non-collinear vectors or independent vectors. Though a\vec a and – b\vec b have same magnitude, but we can’t express a\vec a or – b\vec b in terms of one another. Two non-collinear vectors describe a plane.
  • Co-planar Vectors: Two parallel vectors or non-collinear vectors are always co-planar to one another. Usually, more than two vectors if they lie within the same plane, they are called co-planar vectors. In the figure below, va\vec a, b\vec b and c\vec c are all co-planar as they lie on the same plane.

Important Fundamental Theorems of Vectors

In Two Dimensions:

If there are two non-zero or non-collinear vectors p\vec p and q\vec q, then any vector x\vec x which lies in the plane of p\vec p and q\vec q can be written as a linear combination of p\vec p and q\vec q. We can also write this as there exists L and M ϵ R such that L *p\vec p + M * q\vec q = x\vec x.

This also proves that if L1p+M1q=L2p+M2qL_1 \vec p + M_1 \vec q = L_2 \vec p + M_2 \vec q exists, then we can write

L_1=L_2 and M_1=M_2

 

In Three Dimensions:

If there are three non-zero or non-collinear vectors p,qandr\vec p, \vec q and \vec r then any vector x\vec x which lies in the plane of p,qandr\vec p, \vec q and \vec r can be written as a linear combination of p,qandr\vec p, \vec q and \vec r We can also write this as there exists L, M ad N ϵ R such that

Lp+Mq+Nr=xL \vec p + M \vec q + N \vec r = \vec x.

This also proves that if L1p+M1q+N1r=L2p+M2q+N2rL_1 \vec p + M_1 \vec q + N_1 \vec r = L_2 \vec p + M_2 \vec q + N_2 \vec r exists, then we can write L_1 = L_2, M_1 = M_2 and N_1 = N_2

Linear Combination:

A vector x\vec x is said to be a linear combination of vectors a1,a2,a3,.,an\vec {a_1}, \vec {a_2}, \vec {a_3},…….,\vec {a_n} in a way that there exists scalar m1,m2,m3,.,mn m_1, m_2, m_3,……., m_n and we can write x=a1m1+a2m2+a3m3++anmn\vec x = \vec{a_1} m_1+ \vec{a_2} m_2+ \vec{a_3} m_3+…+\vec{a_n} m_n

Linear Independence:

A system of vectors a1,a2,a3,.,an\vec {a_1}, \vec {a_2}, \vec {a_3},…….,\vec {a_n} are said to be linearly independent

If x=a1m1+a2m2+a3m3++anmn=0\vec x = \vec{a_1} m_1+ \vec{a_2} m_2+ \vec{a_3} m_3+…+\vec{a_n} m_n = 0 which means that

m_1=m_2=m_3…=m_n=0

Condition:

  • A pair of non-collinear vectors is linearly independent.
  • A triad of non-coplanar vector is linearly independent.

Linear Dependence:

A set of vectors a1,a2,a3,.,an\vec {a_1}, \vec {a_2}, \vec {a_3},…….,\vec {a_n} is said to be linearly dependent such that scalars m1,m2,m3,.,mn m_1, m_2, m_3,……., m_n exists such that all of the scalars are not equal to zero and

a1m1+a2m2+a3m3++anmn=0\vec{a_1} m_1+ \vec{a_2} m_2+ \vec{a_3} m_3+…+\vec{a_n} m_n = 0

Condition:

  • A pair of collinear vectors is linearly independent.
  • A triad of coplanar vector is linearly independent.

Theorem 1:

If aandb\vec a and \vec b be two non-collinear vectors, then every vector i\vec i which is co-planar with aandb\vec a and \vec b can be expressed in one and only one combination in the form of xa+yb=i x \vec a + y \vec b= \vec i where x and y are scalar components of the respective vectors.

If two vectors are perpendicular to each other, then the vectors can be supposed to be drawn along the X-axis and Y-axis respectively. If the unit vectors along the two vectors is represented by i^\hat{i} respectively, then we can write that

t=xi+yj \vec t = x \vec i + y \vec j

Theorem 2:

If a,bandc\vec a, \vec b and \vec c are non-coplanar vectors, then any vector \vec{i}[\latex] can be uniquely expressed as a linear combination i=xa+yb+zc \vec i = x \vec a + y \vec b + z \vec c where x, y and z are scalar components of the respective vectors.

Theorem 3:

If vectors a=a1i^+a2j^+a3k^ \vec a = a_1 \hat i + a_2 \hat j + a_3 \hat k b=b1i^+b2j^+b3k^ \vec b = b_1 \hat i + b_2 \hat j + b_3 \hat k c=c1i^+c2j^+a3k^ \vec c = c_1 \hat i + c_2 \hat j + a_3 \hat k

are co-planar, then

a1a2a3b1b2b3c1c2c3\begin{vmatrix} a_1 &a_2 &a_3 \\ b_1& b_2 &b_3 \\ c_1& c_2 &c_3 \end{vmatrix}

Notes:

  1. Three points with position vectors a,bandc\vec a, \vec b and \vec c are collinear if and only if there exist scalars like x, y and z all of which is not equal to zero such that
xa+yb+zc=0x \vec{a} + y \vec{b} + z \vec{c} = 0 and x + y + z = 0

  1. Four points with position vectors a,bcandd\vec a, \vec b \vec c and \vec d are coplanar if and only if there exist scalars like x, y, z and w (sum of any two is not equal to zero) such that
xa+yb+zc+wd=0x \vec{a} + y \vec{b} + z \vec{c} + w \vec{d} = 0 and x + y + z + w = 0

Questions:

Question 1: Points P (p \vec{p}), Q (q \vec{q}), R (r \vec{r}) and S (s \vec{s}) are related as ap+bq+cr+ds=0a \vec{p} + b \vec{q}+ c \vec{r} + d \vec{s}=0 ), and a + b + c + d = 0 (where a, b, c, d are scalars and sum of any two is not zero). Prove that if P, Q, R and S are concyclic then

abpq2=cdrs2|ab||\vec p – \vec q|^2 = |cd| |\vec r – \vec s|^2

Answer:

From the question given, it is understood that P, Q, R and S are coplanar.

Now they are concyclic also.

So, we can write

PO x QO = RO x SO

Or

Fundamental Theorem of Vectors Example

Question 2: Vectors vecpvec p and vecqvec q are non-collinear. Find for what value of y vector vecw=(x2)p+qvec w = (x – 2)\vec p + \vec q and a=(2x+1)pq\vec a = (2x + 1)\vec p – \vec q are collinear?

Answer: