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

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Presentation on theme: "Vectors in 2 and 3 dimensions Definition: A scalar has magnitude A vector has magnitude and direction."— Presentation transcript:

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

2 Representing vectors A B AB= 6 3 6 3 a a = = 6i + 3j Unit Vectors

3 The magnitude of a vector A B 6 3 a

4 3 Dimensions In 3 dimensions, we consider the x, y and z axis. As unit vectors, we use i, j, k E.g.

5 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:

6 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

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

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

9 Equal vectors Two vectors are equal if they have the same magnitude and direction. All of the following vectors are equal: They are the same length and parallel. abcdefg

10 The negative of a vector Here is the vectorAB= 5 2 a A B Suppose the arrow went in the opposite direction: A B We can describe this vector as: BA–a–a –5 –2 or

11 Multiplying vectors by scalars The vector 2a has the same direction but is twice as long. a = 3 2 2a = 6 4 a 2a2a

12 Adding vectors Ifa= 5 3 and b= 3 –2 Find a + b We can represent this addition in the following diagram: a b a + b a + b = 8 1

13 Subtracting vectors andb= –2 3 a= 4 4 a b a – b a – b = 6 1 –b a

14 Subtracting vectors andb= –2 3 a= 4 4 a a – 2b a – 2b = 4 4 –2b b -2 3 4 4 4 -6 8 -2 - 2 = +=

15 O K J IH G F ED C LM b a

16 Vectors on a tangram A tangram is an ancient Chinese puzzle in which a square ABCD is divided as follows: A B C D F E G H I J Suppose, AE= aandAF= b Write the following in terms of a and b. FC= b a 3b3b HJ=–a–a IG=a – 2b CB=2a – 4b HI=b – a HD=2b – 3a

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

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

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

20 Additional points to note: A position vector is one that starts from the origin, such as If one vector is a multiple of the other, the vectors are parallel


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