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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 :



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.

Vector Addition - Analytical Method

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


` \ \ \ \ \ \ =(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.


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



Now we have,



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



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)


`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.


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`