We can now resolve a vector A in terms of component vectors that lie along unit vectors î and ĵ. Consider a vector A that lies in x – y plane as shown in figure. We draw lines from the head of A perpendicular to the coordinate axes as in figure and get vectors A_{1} and A_{2} such that A_{1 + }A_{2 = }A since A_{1} is parallel to î and A_{2} is parallel to ĵ. we have:

A_{1} = A_{x} î, A_{2}= A_{y} ĵ

Where A_{x} and A_{y }are real numbers.

Thus, A = A_{x} î + A_{y} ĵ

This is represented in figure. 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} î is a vector, and so is A_{y} ĵ. Using simple trigonometry, we can express A_{x} and A_{y} in terms of the magnitude of A and the angle θ it makes with the axis:

A_{x} = A cos θ

A_{y }= A sin θ

As is clear from A_{y }= A sin θ a component of a vector can be positive, negative or zero depending on the value of θ.

Now, we have two ways to specify a vector A 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 and θ are given, A_{x} and A_{y} can be obtained using A_{y }= A sin θ. If Ax and A_{y} are given, A and θ can be obtained as follows:

A^{2}_{x} + A^{2}_{y }= A^{2} cos^{2} θ + A^{2} sin^{2} θ

A^{2}_{x} + A^{2}_{y }= A^{2}

Or A = √ (Aₓ² + A_{y}²)

And tanθ = A_{y}/Aₓ, θ = tan⁻¹ A_{y}/AₓA_{x} = A cosα, A_{y} = A cos β, A_{z }= A cos ɣ

In general, we have

A = A_{x} î_{ + }A_{y} ĵ + A_{z }k̂

The magnitude of vector A is A = √ (Aₓ² + A_{y}² + A_{z}²)

A position vector r can be expressed as r = x î + y ĵ + zk̂.

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