Physics RESOLUTION OF VECTORS, VECTOR ADDITION - ANALYTICAL METHOD
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Topics Covered

• Resolution of Vectors
• Vector Addition - Analytical Method
• Unit Vector

Resolution of Vectors

Let veca and vecb be any two non-zero vectors in a plane with different directions and let vecA be another vector in the same plane(Fig.).

vecA can be expressed as a sum of two vectors – one obtained by multiplying veca by a real number and the other obtained by multiplying vecb by another real number. Then, we have

vecA = vec(OP) = vec(OQ) + vec(QP)

But since vec(OQ) is parallel to veca, and vec(QP) is parallel to vecb, we can write

vec(OQ)=lamdaveca, and vec(QP)=muvecb

where lamda and mu are real numbers.

:. vecA=lamdaveca+muvecb

We say that vecA has been resolved into two component vectors lamda veca and mu vecb along veca and vecb 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.

\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Unit Vector")}

A unit vector is a vector of unit magnitude and points in a particular direction.

Unit vectors along the x-, y and z-axes of a rectangular coordinate system are denoted by hati, hatj and hatk, respectively, as shown in Fig. (c).

\color{red} ✍️ It has no dimension and unit.

\color{red} ✍️ It is used to specify a direction only.

\color{red} ✍️ Since these are unit vectors, we have
|hati|= |hatj| =|hatk|=1

These unit vectors are perpendicular to each other.

\color{red} ✍️ If we multiply a unit vector, say vecn by a scalar, the result is a vector.
lamda hatn=veclamda. In general, a vector vecA can be written as vecA = |vecA| hatn
where hatn is a unit vector along vecA.

\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Resolution of Vectors in 2-Dimensions")}

We can now resolve a vector vecA in terms of component vectors that lie along unit vectors hati and hatj. Consider a vector vecA that lies in x-y plane as shown in Fig. (d). We draw lines from the head of vecA perpendicular to the coordinate axes as in Fig. (d), and get vectors vecA_1 and vecA_2.

vecA_1 + vecA_2 = vecA

we have, vecA_1=A_xhati, vecA_2=A_yhatj

where A_x and A_y are real numbers.

Thus, vecA=A_xhati+A_yhatj [This is represented in Fig. (e).]

A_x = x- component of vector vecA

A_y = y- component of vector vecA

\color{red} ✍️ A_x is itself not a vector, but A_xhati is a vector, and so is A_y hatj.

\color{red} ✍️ Magnitude, A_x=Acos theta, A_y=Asin theta
where theta= vecA makes an angle with x-axis.

\color{red} ✍️ 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 vecA 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_x and A_y are given, A and θ can be obtained as follows :

A_x^2+A_y^2=A^2cos^2theta+A^2sin^2theta=A^2

A=sqrt(A_x^2+A_y^2)

and tan theta=(A_y)/(A_x), theta=tan^(-1)[(A_y)/(A_x)]

\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Resolution of Vectors in 3-Dimensions")}

So far we have considered a vector lying in an x-y plane. The same procedure can be used to resolve a general vector vecA into three components along x-, y-, and z-axes in three dimensions.
If alpha,beta and gamma are the angles between vecA and the x-, y-, and z-axes, respectively Fig. (f), we have
A_x = A cos alpha, A_y = A cos beta, A_z =Acos gamma

\color{red} ✍️ In general, we have vecA=A_xhati + A_y hatj + A_z hatk

\color{red} ✍️ The magnitude of vector vecA is A=sqrt(A_x^2 +A_y^2 +A_z^2)

\color{red} ✍️ A position vector vecr can be expressed as vecr=xhati+yhatj+zhatk

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

Consider two vectors vecA and vecB in x-y plane with components A_x, A_y and B_x, B_y :

color{green}{vecA=A_x hati +A_y hatj}

color{green}{vecB=B_x hati +B_y hatj}

Let vecR be their sum. We have

color(blue)(vecR=vecA+vecB)

 \ \ \ \ \ \ =(A_x hati +A_y hatj)+(B_x hati +B_y hatj)

Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors.

color{blue}{vecR=(A_x+B_x)hati+(A_y+B_y)hatj}

Since vecR=R_xhati+R_yhatj

we have, R_x=A_x+B_x, R_y=A_y+B_y

Thus, each component of the resultant vector vecR is the sum of the corresponding components of vecA and vecB.

color{red}{ul"In three dimensions"}, we have
color(green)(vecA=A_xhati+A_yhatj+A_zhatk)

color(green)(vecB=B_xhati+B_yhatj+B_zhatk)

color(green)(vecR=vecA+vecB=R_xhati+R_yhatj+R_zhatk)

Now we have,
R_x=A_x+B_x

R_y=A_y+B_y

R_z=A_z+B_z

This method can be extended to addition and subtraction of any number of vectors. For example, if vectors veca, vecb and vec c are given as
veca=a_xhati+a_yhatj+a_zhatk

vecb=b_xhati+b_yhatj+b_zhatk

vecc=c_xhati+c_yhatj+c_zhatk

then, a vector color(blue)(vecT = veca + vecb – vec c) has components :
T_x =a_x+ b_x -c_x

T_y =a_y+ b_y -c_y

T_z=a_z+ b_z -c_z

Sum of vectors from their magnitude and angle between them

Let OP and OQ represent the two vectors A and B making an angle θ (Fig.). Then, using the parallelogram method of vector addition, OS represents the resultant vector R :

R = A+ B

SN is normal to OP and P M is normal to OS.
From the geometry of the figure,
OS^2 = ON^2+SN^2

but  ON = OP + PN = A + B cos θ

OS^2 = (A + B cos θ)^2 + (B sin θ)^2

or R^2 = A^2+B^2+2AB costheta

R = sqrt(A^2+B^2+2AB costheta) ......(1)

In Δ OSN, SN = OS sinα = R sinα, and in Δ PSN, SN = PS sin θ = B sin θ

Therefore, R sin α = B sin θ

or R/(sintheta) = B/(sinalpha) ...........(2)

Similarly,

PM = A sin α = B sin β

or A/(sin beta) = B/(sinalpha) ........(3)

Combining Eqs. (2) and (3), we get

R/(sintheta) = A/(sinbeta) = B/(sin alpha) ......(4)

Using Eq. (4), we get:

sinalpha = B/R sin theta ........(5)

where R is given by Eq. (1).

or tanalpha = (SN)/(OP +PN) = (B sin thea)/(A+B costheta) .........(6)

Equation (1) gives the magnitude of the resultant and Eqs. (5) and (6) its direction. Equation (1) is known as the Law of cosines
and Eq. (4) as the Law of sines.
Q 3055180064

A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.

Solution:

The vector v_b representing the velocity of the motorboat and the vector v_c representing the water current are shown in Fig. in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.

We can obtain the magnitude of R using the Law of cosine

R = sqrt(v_b^2+v_c^2+2v_bv_c cos120^0)

 = sqrt(25^2+10^2+2xx25xx10(-1/2)) approx 22 km//h

To obtain the direction, we apply the Law of sines

R/(sintheta) = v_c/(sinphi) or sinphi = v_c/R sintheta

 = (10xxsin 120^0)/(21.8) = ( 10 sqrt3)/(2xx21.8) = 0.397

phi equiv 23.4^0