# Vectors in 2 and 3 dimensions

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Vectors in 2 and 3 dimensions
Definition: A scalar has magnitude A vector has magnitude and direction

Representing vectors A B a 3 6 AB = 6 3 a = = 6i + 3j Unit Vectors
Explain that, as with coordinates, movements to the right are positive, movements to the left are negative, movements up are positive and movements down are negative. Link this to work done on transformations. a = = 6i + 3j Unit Vectors

The magnitude of a vector
B a 3 6

3 Dimensions In 3 dimensions, we consider the x, y and z axis.
As unit vectors, we use i, j, k E.g.

Magnitude of a 3D vector The formula for working out the magnitude (length) of a 3D vector is very similar to that for a 2D vector:

Finding Vectors of Given Lengths
Finding Unit Vectors A unit vector in the direction of v can be found as: A vector of length a in the direction of a vector, v, can be found as: Finding Vectors of Given Lengths

Finding Unit Vectors Example: Find the unit vector in the direction of 5i -2j + 4k

Finding Unit Vectors Example: Find the vector of length in the direction of 5i -2j + 4k

Equal vectors Two vectors are equal if they have the same magnitude and direction. All of the following vectors are equal: d g a e c f b Point out equal vectors can be anywhere on the grid. They are the same length and parallel.

The negative of a vector
B Here is the vector AB = 5 2 A B Suppose the arrow went in the opposite direction: The negative of a vector is sometimes called its inverse. Recall that inverse translations map objects back to there starting point. Applying a vector followed by its inverse results in the zero vector, 0. We can describe this vector as: BA –a –5 –2 or

Multiplying vectors by scalars
The vector 2a has the same direction but is twice as long. 2a a a = 3 2 Point out that when the line is twice as long the horizontal and vertical components are doubled. 2a = 6 4

Adding vectors If a = 5 3 and b –2 Find a + b
We can represent this addition in the following diagram: Explain that adding these two vectors is like moving right 5 and up 3 and then moving right 3 and down 2. The net effect is a movement right 8 and up 1. Point out that we can add the horizontal components together to get the horizontal component of the resultant vector (5 + 3 = 8) and we can add the vertical components together to get the vertical component of the resultant vector (3 + –2 = 1). In the vector diagram the start of vector b is placed at the end of vector a. The resultant vector, a + b, goes from the start of a to the end of b. b a a + b = 8 1 a + b

Subtracting vectors and b = –2 3 a 4 a b –b a –b a – b a – b = 6 1
Explain that to draw a diagram of a – b we draw vector a followed by vector –b. The resultant vector a – b goes from the beginning of a to the end of –b. Establish again that subtracting the horizontal components gives 4 – –2 = 6 and subtracting the vertical components gives 4 – 3 = 1. a – b a – b = 6 1

Subtracting vectors and b = –2 3 a 4 a –2b b a – 2b 4 -2 3 4 4 -6 8 -2
Explain that to draw a diagram of a – b we draw vector a followed by vector –b. The resultant vector a – b goes from the beginning of a to the end of –b. Establish again that subtracting the horizontal components gives 4 – –2 = 6 and subtracting the vertical components gives 4 – 3 = 1. a – 2b 4 -2 3 4 4 -6 8 -2 a – 2b = = + =

J K L M F G H I b O C D E a

Write the following in terms of a and b.
Vectors on a tangram A tangram is an ancient Chinese puzzle in which a square ABCD is divided as follows: D C Suppose, AE = a and AF = b F E G H I J Write the following in terms of a and b. FC = 3b CB = 2a – 4b Explain how the square is divided to make a tangram. For example, tell pupils that H is the mid-point of CB and E is the midpoint of AB. Explain that more difficult vectors can be broken up into smaller vectors and added together. For example, HD = HI + IC + CD = (b – a) + b + –2a = 2b – 3a. Ask how we can use these vectors to show that IG and CB are parallel and that CB is twice the length of IG. Ask pupils to give other vectors on this grid in terms of a and b. b HJ = –a HI = b – a A a B IG = a – 2b HD = 2b – 3a

C is ¼ of the way along OB B OC = b BC = C O

C divides the line OB in the ratio 1:5
OC = s BC = C O

C divides the line AB in the ratio 1:2
= -a + b AC = 1/3 (–a + b) b OC = C A O a