[Physics Class Notes] on Resolution of Vector Rectangular Components Pdf for Exam

Resolution of a Vector in a Plane

All Physical quantities like force, momentum, velocity, acceleration are all vector quantities because they have both magnitude and direction. We represent the vector as an arrow-headed line, where the tip of the arrow is the head and the line is the tail.

Let’s suppose there are two paths, viz: A and B, where A and B are horizontal and vertical components of a vector, respectively. So, the displacement can be calculated by using a Pythagoras theorem from the following diagram:

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[sqrt{3^{2}}] + [sqrt{4^{2}}] = [sqrt{9}] + [sqrt{16}] = [sqrt{25}] = 5    

So,  5 m is displacement.

Now, let’s discuss what is the horizontal and vertical component.

Horizontal Component Definition

In science, we define the horizontal component of a force as the part of the force that moves directly in a line parallel to the horizontal axis. 

Let’s suppose that you kick a football, so now, the force of the kick can be divided into a horizontal component, which is moving the football parallel to the ground, and a vertical component that moves the football at a right angle to the surface/ground.

Vertical Component Definition

We define the vertical component as that part or a component of a vector that lies perpendicular to a horizontal or level plane.

Resolution of a Vector

Resolution of a vector is the splitting of a single vector into two or more vectors in different directions which together produce a similar effect as is produced by a single vector itself. The vectors formed after splitting are called component vectors.

Let’s understand this with the following diagram:

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Here, 

OB[^{rightarrow}] = ax =  vector along the x-axis (It is the horizontal component formula)

OD[^{rightarrow}] = ay = vector along the y-axis (It s the vertical component formula)

From here, we obtained the horizontal and vertical components of a vector, which is a vector a[^{rightarrow}].

From the triangle law of addition, we can use the formula as:

OC[^{rightarrow}] = OB[^{rightarrow}] + OD[^{rightarrow}]

a[^{rightarrow}] = ax +  ay….(1)

Here, we can see that OCB is right-angled, so using the formula of the trigonometric function, we get the angular components along the x and y-axis, respectively:

Since

OB/OC = Cos

OB = OC Cos

So, 

ax = a[^{rightarrow}] Cos….(2)

Similarly, 

BC/OC = Sin

ay = a[^{rightarrow}] Sin….(3)

Now, eq (3) ÷ eq (2), we get the tangent of component, which is given by:

(a[^{rightarrow}] SinΘ)/a[^{rightarrow}] CosΘ() = ay/ ax 

So,

tan = BC/OB = ay/ ax ….(4)

Rectangular Components of Vectors in Three Dimensions

We define rectangular components of vectors in Three Dimensions in the following manner:

If the coordinates of a point P, i.e., x, y, and z, the vector joining point P to the origin is called the position vector. The position vector of point P is equal to the sum of these coordinates, which is given by:

         x + y + z

Rectangular Components 

Rectangular components of a vector in three dimensions can be better understood by going through the following context:

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Let’s suppose that vector A[^{rightarrow}] is presented by the vector OR[^{rightarrow}]. Now, taking O as the origin and construct a rectangular parallelopiped with its three edges along with the three rectangular axes, viz: X, Y, and Z. Here, we can notice that A[^{rightarrow}] represents the diagonal of the rectangular parallelopiped whose intercepts are the ax,  ax,  and ax, respectively. We call these intercepts the three rectangular components of A[^{rightarrow}]. 

Now, using the triangular law of vector addition, we have:

OR[^{rightarrow}]  + OT[^{rightarrow}]  + TR[^{rightarrow}] 

Using the parallel law of vector addition, we have:

OT[^{rightarrow}] = OS[^{rightarrow}] + OP[^{rightarrow}]

OR[^{rightarrow}] = (OS[^{rightarrow}] + OP[^{rightarrow}]) + TR[^{rightarrow}] ……(5)

Here, one must notice that TR[^{rightarrow}] = OQ[^{rightarrow}]. So, rewriting equation (5) in the following manner:

OR[^{rightarrow}] = (OS[^{rightarrow}] + OP[^{rightarrow}]) + OQ[^{rightarrow}]

Or,

A[^{rightarrow}] = A[^{rightarrow}]z + A[^{rightarrow}]x + A[^{rightarrow}]y = A[^{rightarrow}]x + A[^{rightarrow}]y + A[^{rightarrow}]z ……(6)

Therefore,

A[^{rightarrow}] = Ai[^{rightarrow}]x + Aj[^{rightarrow}]y + Ak[^{rightarrow}]z

Also, 

OR² = OT² + TR²

OP² + OS² + TR²

Now,

A² = A²x + A²y + A²z ……..(7)

Resolution of Rectangular Vectors in Three Dimensions in their Direction Cosines 

Now, we can restate equation (6) in the following manner:

A = [sqrt{A^{2}x + A^{2}y + A^{2}z}]

If α, β, and γ are the angles which the vector A[^{rightarrow}] makes with the X, Y, and Z-axis, respectively, then we have:

Cos α = Ax[^{rightarrow}] / A[^{rightarrow}] ⇒ Ax[^{rightarrow}] = A[^{rightarrow}] Cos α …..(a)

Cos β = A[^{rightarrow}]y / A[^{rightarrow}] ⇒ A[^{rightarrow}]y = A[^{rightarrow}] Cos β ….(b)

Cos γ = Az[^{rightarrow}] / A[^{rightarrow}] ⇒ Az[^{rightarrow}] = A[^{rightarrow}] Cos γ …..(c)

We must note that Cos α,  Cos β, and Cos γ are direction cosines of vectors Ax[^{rightarrow}], Ay[^{rightarrow}], and Az[^{rightarrow}], respectively.

Now, putting the values of equations (a), (b), and © in the equation (7), we get:

A² = A² Cos²α + A²Cos² β + A² Cos² γ …….(8)

So, we get the equation as:

Cos² α + Cos² β + Cos² γ = 1 

Here, we conclude that the squares of the direction cosines of three vectors are always constant, i.e., unity.