1. Vector
geometry

2. Starter: name four types of transformation of a shape
Vectors: Geometry
KUS objectives
BAT solve problems with vector geometry; understand the basics of vector geometry including addition and scalar multiplication
Starter: name four types of transformation of a shape
Reflection over a given line
Rotation cw or acw around a given point by an angle
Translation by a vector
Enlargement by a scale factor from a given point

3. Direction and Magnitude
Notes 1
Direction and Magnitude
Scalar
Vector
The distance from P to Q is 100m P Q
From P to Q you go 100m north
A scalar quantity has only a magnitude (size) A vector quantity has both a magnitude and a direction N
60°
Scalar
A ship is sailing at 12km/h
Vector
A ship is sailing at 12km/h on a bearing of 060°

4. Alternatively, single letters can be used…
Notes 2 Q
Equal vectors have the same magnitude and direction S P R
A Common way of showing vectors is using the letters with an arrow above
PQ = RS a
Alternatively, single letters can be used… b

5. Use Pythagoras’ Theorem
Notes 3 7 4 p
Use arctan
North
The direction value of a vector is often given
as a bearing from North
Example 𝑝= 7 −4
Then the bearing from North is 90+ arctan =120° (3𝑠𝑓)
The modulus value of a vector is another name for its magnitude Eg) The modulus of the Vector a is |a| The modulus of the vector PQ is |PQ| Example The vector a is directed due east and |a| = 12. Vector b is directed due south and |b| = 5. Find |a + b| a b
a + b
Use Pythagoras’ Theorem

6. Q
Notes 4 a
Adding the vectors PQ and QP gives a Vector result of 0 Vectors of the same size but in opposite directions have opposite signs (eg) + or - Q P -a P A C B
The triangle law for vectors 𝐴𝐵 + 𝐵𝐶 = 𝐴𝐶
Also
𝒂+𝒃=𝒃+𝒂
The resultant vector is the vector sum of two or more vectors a b
a+b

7. WB 1 P Q R T S a b c d
In the diagram opposite, find the following vectors in terms of a, b, c and d 𝑃𝑆 𝑅𝑃 𝑃𝑇 𝑇𝑆
-c + a Or c - a
-b + a Or a - b
-a + b + d Or b + d - a
-d - b + c Or c – b - d

8. The second Vector is a multiple of the first, so they are parallel
WB2 parallel vectors
Any vector parallel to a may be written as λa, where λ (lamda) is a non-zero scalar (ie - represents a number…)
a) Show that the vectors 6a + 8b and 9a + 12b are parallel…
9𝑎+12𝑏= 𝑎+8𝑏
The second Vector is a multiple of the first, so they are parallel
b) Show that the vectors 5b - 2a and 14a - 35b are parallel…
−2𝑎+5𝑏=− 𝑎−35𝑏

9. = 3 5 𝐴𝐵 so PQ and AB are parallel
WB3 In triangle OAB 𝑂𝐴 =𝑎 , 𝑂𝐵 =𝑏, P divides OA in the ratio 3:2, and Q divides OB in the ratio 3:2
Show that PQ is parallel to AB
Given that AB is 10 cm in length, find the length of PQ
a) 𝑃𝑄 = 𝑃𝑂 + 𝑂𝑄
=− 3 5 𝑎+ 3 5 𝑏
= 3 5 −𝑎+𝑏
= 3 5 𝐴𝐵 so PQ and AB are parallel
b) ×10 =6 𝑐𝑚

10. Triangle Law for vectors
 Two vectors can be added using the ‘Triangle Law’ b a
It is important to note that vector a + b is the single line from the start of a to the end of b.
Vector a + b is NOT the two separate lines!
a + b
Example Draw a diagram to show the vector a + b + c a b c a b c
a + b + c

11. WB 4 Triangle Law for vectors𝑏 𝑎
i) 𝑎 and 𝑏 must be parallel
Careful with diagrams !
Use the sine rule
Q from 2019 exam paper
sin ∅ = sin 30° 𝑚−𝑛 × 𝑚
30° n 𝑚
sin ∅ = sin 30° 6 ×3= 1 4 𝑚 −𝑛
𝑚−𝑛 ∅
30°
∅= arcsin =14.5°
So the angle between is −30−14.5=135.5°

12. SR = 4a and PX = kPR. Find in terms of a, b and k:
WB 5 In the diagram, PQ = 3a, QR = b, P Q R S X 3a 4a b
SR = 4a and PX = kPR. Find in terms
of a, b and k: 𝑃𝑆 𝑃𝑋 𝑆𝑄 𝑆𝑋
= 3a + b – 4a = b – a
= kPR = k(3a + b)
= 4a - b
= -b + a + k(3a + b)
= (3k + 1)a + (k – 1)b
e) Use the fact that X lies on SQ to find the value of k
𝑆𝑋 =𝛾 4a − b
= (3k + 1)a + (k – 1)b
Equate the coefficients 4𝛾=3𝑘+1
− 𝛾=𝑘−1
Solving gives 𝑘= 3 7

13. 𝐵 𝑆 𝐶 𝑁 𝐷 𝑀 a) 𝑀𝑁 = 𝑀𝑂 + 𝑂𝐴 + 𝐴𝑁 𝑀𝑁 =− 1 5 𝑏+𝑎+ 4 5 𝑏 𝑂 𝑇 𝐴
WB 6a OACB is a parallelogram with 𝑂𝐴 =𝑎 and 𝑂𝐵 =𝑏
The points M, S, N and ||T divide OB, BC, CA and AO in the ratio 1:4 respectively
The lines ST and MN intersect at point D
Express 𝑀𝑁 and 𝑆𝑇 in terms of a and b
Show that lines MN and ST bisect each other 𝑀 𝑂 𝐴 𝑁 𝐵 𝐶 𝑆 𝑇 𝐷
a) 𝑀𝑁 = 𝑀𝑂 + 𝑂𝐴 + 𝐴𝑁
𝑀𝑁 =− 1 5 𝑏+𝑎+ 4 5 𝑏
𝑀𝑁 = 𝑎+ 3 5 𝑏
𝑆𝑇 = 𝑆𝐵 + 𝐵𝑂 + 𝑂𝑇
𝑆𝑇 =− 1 5 𝑎+ −𝑏 𝑎
𝑆𝑇 = 3 5 𝑎 − 𝑏

14. WB 6b OACB is a parallelogram with 𝑂𝐴 =𝑎 and 𝑂𝐵 =𝑏
The points M, S, N and ||T divide OB, BC, CA and AO in the ratio 1:4 respectively
The lines ST and MN intersect at point D
Express 𝑀𝑁 and 𝑆𝑇 in terms of a and b
Show that lines MN and ST bisect each other 𝑀 𝑂 𝐴 𝑁 𝐵 𝐶 𝑆 𝑇 𝐷
= 𝛾 𝑎 𝛾 𝑏
b) Let 𝑀𝐷 =𝛾 𝑎+ 3 5 𝑏
𝑆𝐷 =𝜇 3 5 𝑎−𝑏
Then we also can have 𝑀𝐷 = 𝑀𝑆 + 𝑆𝐷
𝑀𝐷 = 4 5 𝑏+ 1 5 𝑎 +𝜇 3 5 𝑎−𝑏 = 𝜇 𝑎 −𝜇 𝑏
Equate the two versions of 𝑀𝐷
𝑀𝐷 = 𝛾 𝑎 𝛾 𝑏 = 𝜇 𝑎 −𝜇 𝑏
Equate the coefficients of a and b
𝛾= 𝜇
Solving simultaneously 𝛾= 𝜇= 1 2
3 5 𝛾= 4 5 −𝜇
So D is the midpoint of both MN and ST
So MN and ST bisect each other

15. comparing coefficients 2𝑠+𝑡 𝑎=5𝑎 𝑠−𝑡 𝑏=−4𝑏
WB 7 Uniqueness
Given that: 5𝑎−4𝑏= 𝑠 2𝑎+𝑏 +𝑡 𝑎−𝑏 find the value of the scalars s and t
5𝑎−4𝑏= 2𝑠+𝑡 𝑎+ 𝑠−𝑡 𝑏
comparing coefficients 2𝑠+𝑡 𝑎=5𝑎
𝑠−𝑡 𝑏=−4𝑏
Effectively, if the two vectors are equal then the coefficients of a and b must also be equal
2𝑠+𝑡 =5
𝑠−𝑡 =−4
Solving simultaneously gives 𝑠= and 𝑡=4 1 3

16. One thing to improve is –
KUS objectives
BAT solve problems with vector geometry; understand the basics of vector geometry including addition and scalar multiplication
self-assess
One thing learned is –
One thing to improve is –

17. END