Logical Class

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. 4.8). 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

OQ = \(\lambda\) a, and QP = µ b(4.7) where \(\lambda\) and µ are real numbers.

fig 4.8.(a)Two non-colinear vectors a and b.(b) Resolving a vector Ain terms of vectors a and b

We say that A has been resolved into two component vectors \(\lambda\) a 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 \(\hat i,\hat j\) and \(\hat k\),respectively, as shown in fig.4.9(a).

Since these are unit vectors, we have

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 n by a scalar, the result is a vector \(\lambda\)=\(\lambda\)\(\hat n\) In general, a vector A can be written as

A = |A| n

where n is a unit vector along A.

We can now resolve a vector A in terms of component vectors that lie along unit vectors \(\hat i\) and \(\hat j\) .Consider a vector A that lies in x-y plane as shown in Fig. 4.9(b). We draw lines from the head of A perpendicular to the coordinate axes as in Fig. 4.9(b), and get vectors A₁ and A₂such that A₁+ A₂ = A. Since A₁ is parallel to \(\hat i\).

and A₂ is parallel to \(\hat j\) , we have

A₁= A_x \(\hat i\) , A₂ = A_y\(\hat j\)

where A_x and A_y are real numbers

Thus, A = A_x \(\hat i\) + A_y \(\hat j\)

This is represented in Fig. 4.9(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\(\hat i\) is a vector, and so is A_y\(\hat j\).Using simple trigonometry, we can express A_x and A_y in terms of the magnitude of A and the angle \(\theta\) it makes

with the x-axis :

As is clear from Eq. (4.13), a component of a vector can be positive, negative or zero depending on the value of \(\theta\).

Now, we have two ways to specify a vector A in a plane. It can be specified by :

1.Its magnitude A and the direction \(\theta\) it makes with the x-axis; or

2.Its components A_x and A_y If A and \(\theta\) are given, A_x and A_y can be obtained using Eq. (4.13). If A_x and A_y are given, A and \(\theta\) can be obtained as follows:

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGbb % Waa0baaSqaaiaadIhaaeaacaaIYaaaaOGaey4kaSIaamyqamaaDaaa % leaacaWG5baabaGaaGOmaaaakiabg2da9iaadgeadaahaaWcbeqaai % aaikdaaaGcciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaikdaaaGc % cqaH4oqCcqGHRaWkcaWGbbWaaWbaaSqabeaacaaIYaaaaOGaci4Cai % aacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaeqiUdehabaGaeyyp % a0JaamyqamaaCaaaleqabaGaaGOmaaaaaOqaaiaadgeacqGH9aqpda % GcaaqaaiaadgeadaqhaaWcbaGaamiEaaqaaiaaikdaaaGccqGHRaWk % caWGbbWaa0baaSqaaiaadMhaaeaacaaIYaaaaaqabaaakeaacaWGbb % GaamOBaiaadsgacaaMc8UaaGPaVlGacshacaGGHbGaaiOBaiabeI7a % Xjabg2da9maalaaabaGaamyqamaaBaaaleaacaWG5baabeaaaOqaai % aadgeadaWgaaWcbaGaamiEaaqabaaaaOGaaiilaiabeI7aXjabg2da % 9iGacshacaGGHbGaaiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaakm % aalaaabaGaamyqamaaBaaaleaacaWG5baabeaaaOqaaiaadgeadaWg % aaWcbaGaamiEaaqabaaaaaaaaa!72EC! \begin{array}{l} A_x^2 + A_y^2 = {A^2}{\cos ^2}\theta + {A^2}{\sin ^2}\theta \\ = {A^2}\\ A = \sqrt {A_x^2 + A_y^2} \\ And\,\,\tan \theta = \frac{{{A_y}}}{{{A_x}}},\theta = {\tan ^{ - 1}}\frac{{{A_y}}}{{{A_x}}} \end{array}\)

fig.4.9. (a)Unit vectors \(\hat i,\hat j\) and \(\hat k\), 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.

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 \(\alpha,\beta\) and \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeq4SdCgaaa!379D! \gamma \) are the angles* between A and the x-, y-, and z-axes, respectively [Fig. 4.9(d)], we have

fig.4.9.(d)A vector A resolved into components along x-, y-, and z-axes.

A _x = A cos \(\alpha\), A_y= A cos \(\beta\), A_z=A cos\(\gamma\)

In general, we have

The magnitude of vector A is

A position vector r can be expressed as

where x, y, and z are the components of r along x-, y-, z-axes, respectively

RESOLUTION OF VECTORS