Let a and b be any two non-zero vectors in a plane with different directions and let A be another vector in the same plane(Fig. 1). A can be expressed as a sum of two vectors — one obtained by multiplying a by a real number and the other obtained by multiplying b by another real number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have
A = OP = OQ + QP
But since OQ is parallel to a, and QP is parallel to b, we can write:
OQ = λa, and QP = µb
where λ and µ are real numbers.
Therefore, A = λa + µb
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| Fig.1: (a) Two non-colinear vectors a and b. (b) Resolving a vector A in terms of vectors a and b. |
We say that A has been resolved into two component vectors λa and µb along a and b respectively. Using this method one can resolve a given vector into two component vectors along a set of two vectors – all the three lie in the same plane. It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude. These are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only. Unit vectors along the x-, y- and z-axes of a rectangular coordinate system are denoted by $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$ and $\mathbf{\hat{k}}$, respectively, as shown in Fig. 2(a).
Since these are unit vectors, we have
|$\mathbf{\hat{i}}$| = |$\mathbf{\hat{j}}$| = |$\mathbf{\hat{k}}$|
These unit vectors are perpendicular to each other. In this text, they are printed in bold face with a cap (^) to distinguish them from other vectors. Since we are dealing with motion in two dimensions in this chapter, we require use of only two unit vectors. If we multiply a unit vector, say $\mathbf{\hat{n}}$ by a scalar, the result is a vector
λ = λ$\mathbf{\hat{n}}$.
In general, a vector A can be written as
where $\mathbf{\hat{n}}$ is a unit vector along A.
We can now resolve a vector A in terms of component vectors that lie along unit vectors $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$. Consider a vector A that lies in x-y plane as shown in Fig. 2(b). We draw lines from the head of A perpendicular to the coordinate axes as in Fig. 2(b), and get vectors $\mathbf{A_1}$ and $\mathbf{A_2}$ such that $\mathbf{A_1}$ + $\mathbf{A_2}$ = $\mathbf{A}$. Since $\mathbf{A_1}$ is parallel to $\mathbf{\hat{i}}$ and $\mathbf{A_2}$ is parallel to $\mathbf{\hat{j}}$, we have :
$\mathbf{A_1}$ = $A_1$$\mathbf{\hat{i}}$, $\mathbf{A_2}$ = $A_2$$\mathbf{\hat{j}}$
where $A_x$ and $A_y$ are real numbers.
Thus, $\mathbf{A}=A_1 \mathbf{\hat{i}}+A_2 \mathbf{\hat{j}}$
This is represented in Fig. 2(c). The quantities A$_x$ and A$_y$ are called x-, and y- components of the vector A. Note that A$_x$ is itself not a vector, but $A_x$$\mathbf{\hat{i}}$ is a vector, and so is $A_y$$\mathbf{\hat{i}}$. Using simple trigonometry, we can express A$_x$ and A$_y$ in terms of the magnitude of A and the angle θ it makes with the x-axis:
A$_x$ = A cos θ
A$_y$ = A sin θ
As is clear from Eq. (4.13), a component of a vector can be positive, negative or zero depending on the value of θ. Now, we have two ways to specify a vector A in a plane. It can be specified by :
(i) its magnitude A and the direction θ it makes with the x-axis; or
(ii) its components A$_x$ and A$_y$ If A and θ are given, A$_x$ and A$_y$ can be obtained using Eq. (4.13). If A$_x$ and A$_y$ are given, A and θ can be obtained as follows:
$A_x^2+A_y^2$ = $A^2$ cos$^2$ θ + $A^2$ sin $^2$ θ = $A^2$
Or $A^2=\sqrt{A_x^2+A_y^2}$
And tan θ = $\frac{A_y}{A_x}$, θ = tan$^{-1}$ $\frac{A_y}{A_x}$
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| Fig.2 : (a) Unit vectors $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$ and $\mathbf{\hat{k}}$ lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components A$_x$ and A$_y$ along x-, and y- axes. (c) $\mathbf{A_1}$ and $\mathbf{A_2}$ expressed in terms of $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$. |
So far we have considered a vector lying in an x-y plane. The same procedure can be used to resolve a general vector A into three components along x-, y-, and z-axes in three dimensions. If α, β, and γ are the angles* between A and the x-, y-, and z-axes, respectively [Fig. 2(d)], we have
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| Fig. 2(d) A vector A resolved into components along x-, y-, and z-axes |
A$_x$ = A cos α, A$_y$ = A cos β, A$_z$ = A cos γ
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