**Resolution of Vectors and Rectangular Components**

The process of a splitting a vector is called **resolution of a vector**. In simpler language it would mean, determining the effect of a vector in a particular direction. The parts of the vector obtained after splitting the vector are known as **Components of the Vector**.

**Rectangular components of a vector: **If the components of a given vector are perpendicular to each other, they are called as **Rectangular components**. The figure illustrates a vector \(\overrightarrow{A}\) represented by \(\overrightarrow{OP}\). Through the point, O two mutually perpendicular axis X and Y are drawn. From the point P, two perpendicular, PN and PM are dropped on X and Y axis respectively.The vector \(\overrightarrow{{{A}_{x}}}\) is the resolved part of \(\overrightarrow{A}\) along the X – axis. It also known as the X – component of \(\overrightarrow{A}\) and is the projection of the \(\overrightarrow{A}\) on X- axis. Similarly, \(\overrightarrow{{{A}_{y}}}\) is the resolved part of the \(\overrightarrow{A}\) along the Y – axis, and is therefore, known as the Y – component of \(\overrightarrow{A}\).

Applying the law of triangle of vectors to ONP, \(\overrightarrow{OP}\,\,=\,\,\overrightarrow{ON}\,+\,\overrightarrow{NP}\) or \(\overrightarrow{A}\,=\,\overrightarrow{{{A}_{x}}}\,\,+\,\,\overrightarrow{{{A}_{y}}}\), which also confirm that A_{x}, A_{y} are the components of A.

Moreover, in the right – angled MONP,

\(\cos \,\theta \,\,=\,\,\frac{{{A}_{x}}}{A}\)

⇒ A_{x} = A cosθ … (1)\(\sin \,\theta \,=\,\frac{{{A}_{y}}}{A}\)

⇒ A_{x} = A sinθ … (2)

Squaring and adding equations (1) and (2) we get,

A_{x}² + A_{y}² = A² cos²θ + A² sin²θ = A² (cos²θ + sin²θ)

But, cos²θ + sin²θ = 1

∴ A_{x}² + A_{y}² = A²

⇒ A² = A_{x}² + A_{y}²

\(A\,=\,\sqrt{{{A}_{x}}^{2}\,\,+\,\,{{A}_{y}}^{2}}\)

This equation gives the magnitude of the given vector in terms of the magnitudes of the components of the given vector.

In the figure, the velocity vector \(\overrightarrow{V}\) is represented by the vector \(\overrightarrow{OP}\). Resolving \(\overrightarrow{V}\) into its two rectangular components, we have \(\overrightarrow{V}\,=\,\overrightarrow{{{V}_{x}}}+\overrightarrow{{{V}_{y}}}\). In terms of the unit vectors \(\widehat{i}\), \(\widehat{j}\), \(\overrightarrow{V}\,\,=\,{{V}_{x}}\,\widehat{i}\,\,+\,{{V}_{y}}\widehat{j}\)

Where,

V_{x} = V cosθ, V_{y} = V sinθ and \(\tan \,\theta \,=\,\frac{{{V}_{y}}}{{{V}_{x}}}\).

**How to find the Resolution of Vectors?**

**Problem: **Diagram above shows two forces of magnitude 25N are acting on an object of mass 2kg. Find the acceleration of object, in m/s².**Solution: **Given,

Mass (m) = 2kg

Horizontal component of the forces = 25 cos 45° + 25 cos 45° = 35.36 N

Vertical component of the forces = 25 cos 45° – 25 cos 45° = 0 N

The acceleration of the object can be determined by the equation Force (F) = mass (m) x acceleration (a)

(35.36) = 2 x a

Acceleration (a) = 35.36/3 = 17.68 m/sec²