 # Vectors for Edexcel A-level Maths

1. Position vectors
2. Vector magnitude
3. Vector direction

A position vector is a vector that describes the position of a point relative to the origin. If a point has coordinates (x, y, z), then its position vector is xi + yj + zk relative to the origin.

When the position vectors of two points are known, the vector between the two points can be found by subtracting one position from the vector. Let OA and OB be the position vectors of the points A and B respectively. The vector AB can then be found by moving from A to the origin, ie. −OA, and from the origin to B, ie. OB, so AB = OB − OA. The magnitude of a vector can be calculated using Pythagoras’ theorem. A vector given by (x y) has magnitude √(x² + y²), and a vector given by (x y z) has magnitude √(x² + y² + z²). To obtain the distance between two points, the vector between them can be found and then its magnitude can be calculated to give the required distance. The direction of a vector can be calculated using trigonometric functions to find the angle which the vector makes with the horizontal. For a 2D vector the direction the vector makes with the x-axis can be found by obtaining an arctangent of the x and y coordinate ratio. Once the magnitude and direction of a vector are known, the vector can be expressed in the following form: v = |v|(icosθ + jsinθ), where |v| is the magnitude and θ is the angle it makes with the horizontal. # 1

Calculate the direction of the vector 4i + 7j with respect to x axis.

For a 2D vector the direction the vector makes with the x-axis can be found by obtaining an arctangent of the x and y coordinate ratio.
tanθ = 7/4
θ = arctan(7/4) = 60.3° anticlockwise

60.3° anticlockwise # 2

Express the given vector in magnitude and direction form.

Let (2, 8) be A and (11, 2) be B.
OA = 2i + 8j
OB = 11i + 2j
AB = OB − OA
AB = 11i + 2j − (2i + 8j) = 9i − 6j

tanθ = 6/9
θ = arctan(6/9) = 33.7° clockwise (to 3sf)

The magnitude can be calculated using Pythagoras’ theorem.
Magnitude = √(9² + (−6)²) = 3√13

v = |v|(cosθi + sinθj)
v = 3√13(cos(−33.7°)i + sin(−33.7°)j)

3√13(cos(−33.7°)i + sin(−33.7°)j) # 3

Given that C is the midpoint of the vector AB, find the position vector of C.

OA = 2i + 10j
OB = 12i + 3j
AB = OB − OA
AB = 12i + 3j − (2i + 10j)
AB = 10i − 7j
AC = 0.5AB = 5i − 3.5j
OC = OA + AC
OC = 2i + 10j + 5i − 3.5j = 7i + 6.5j

7i + 6.5j End of page