2.2 Finding a Vector from its Components (Vector Composition) | Vectors | Physics 11

Vector composition is the process of determining a single vector from its perpendicular components. This is the reverse of vector resolution, which is covered in Rectangular Components of a Vector→. Vector composition allows us to combine horizontal (x-component) and vertical (y-component) influences to find the true magnitude and direction of the resultant vector. This skill is essential for analyzing forces, velocities, and displacements in two dimensions.

Figure 1: Vector composition showing components and resultant vector

Given the perpendicular components of a vector, typically $A_{x}$ (the x-component) and $A_{y}$ (the y-component), we can find the original vector $A$ by determining its magnitude and direction.

Visualizing the Vector

The components $A_{x}$ and $A_{y}$ can be visualized as the legs of a right-angled triangle.

The x-component, $A_{x}$ , is the base of the triangle.
The y-component, $A_{y}$ , is the height of the triangle.
The resultant vector, $A$ , is the hypotenuse, connecting the starting point of $A_{x}$ to the endpoint of $A_{y}$ .

1. Calculating the Magnitude of the Vector

The magnitude of a vector is its length. Since the components form a right-angled triangle, we can use the Pythagorean theorem to find the magnitude ( $A$ ).

Formula: $A = A_{x}^{2} + A_{y}^{2}$

Explanation: The square of the magnitude of the resultant vector (the hypotenuse) equals the sum of the squares of its components (the two legs).

Example: A force has a horizontal component $F_{x} = 3.0 N$ and a vertical component $F_{y} = 4.0 N$ . Find its magnitude. $F = (3.0)^{2} + (4.0)^{2} = 9.0 + 16.0 = 25.0 = 5.0 N$

In three dimensions, the vector is represented as: $A = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$ and its magnitude extends to: $A = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}$

2. Determining the Direction of the Vector

The direction of a vector is the angle it makes with a reference axis, typically the positive x-axis. We can find this angle using the inverse tangent.

Formula: $θ = tan^{- 1} (\frac{A _{y}}{A _{x}})$

Explanation: The tangent of angle $θ$ in a right-angled triangle is the ratio of the opposite side ( $A_{y}$ ) to the adjacent side ( $A_{x}$ ).

Important Note on Quadrants: The standard $tan^{- 1}$ function returns an angle only between $- 9 0^{\circ}$ and $+ 9 0^{\circ}$ . To find the correct angle in the full $36 0^{\circ}$ system, you must consider the signs of $A_{x}$ and $A_{y}$ .

Let $ϕ = tan^{- 1} \frac{A _{y}}{A _{x}}$ be the reference angle.

Quadrant	Sign of $A_{x}$	Sign of $A_{y}$	Direction Angle ( $θ$ )
I	+	+	$θ = ϕ$
II	-	+	$θ = 18 0^{\circ} - ϕ$
III	-	-	$θ = 18 0^{\circ} + ϕ$
IV	+	-	$θ = 36 0^{\circ} - ϕ$

Example (Quadrant II): A displacement vector has components $D_{x} = - 3.0 m$ and $D_{y} = 4.0 m$ . Find its direction.

Find the reference angle $ϕ$ : $ϕ = tan^{- 1} (\frac{4.0}{- 3.0}) = tan^{- 1} (1.33) \approx 53. 1^{\circ}$
Determine the correct angle $θ$ : Since $A_{x} < 0$ and $A_{y} > 0$ , the vector is in Quadrant II. $θ = 18 0^{\circ} - 53. 1^{\circ} = 126. 9^{\circ}$

The direction is $126. 9^{\circ}$ counter-clockwise from the positive x-axis.

Special Case: Equal Components

When $A_{x} = A_{y}$ (both positive), the vector makes an angle of $4 5^{\circ}$ with the x-axis.

Quadrant

Sign of

A_{x}

Sign of

A_{y}

Direction Angle (

θ

)

θ = ϕ

θ = 18 0^{\circ} - ϕ

III

θ = 18 0^{\circ} + ϕ

θ = 36 0^{\circ} - ϕ