Ideas for a Further Mathematics classroom
WHY BOTHER WITH VECTOR EQUATIONS
Starter…
Particle A is projected from the position 3, −1,2 with
velocity −1,6,2 .
At the same instant, particle B is projected from the position
5,5,4 with velocity −2,3,1 .
Do the particles collide?
Vectors make things easier…
1 2
𝐫𝐴 = 𝐬𝐴 + 𝐮𝐴 𝑡 + 𝐚𝑡
2
1 2
𝐫𝐵 = 𝐬𝐵 + 𝐮𝐵 𝑡 + 𝐚𝑡
2
1 2
1 2
𝐬𝐴 + 𝐮𝐴 𝑡 + 𝐚𝑡 = 𝐬𝐵 + 𝐮𝐵 𝑡 + 𝐚𝑡
2
2
𝐬𝐴 + 𝐮𝐴 𝑡 = 𝐬𝐵 + 𝐮𝐵 𝑡
𝐬𝐴 − 𝐬𝐵 = 𝐮𝐵 𝑡 − 𝐮𝐴 𝑡
𝐬𝐴 − 𝐬𝐵 = −𝑡 𝐮𝐴 − 𝐮𝐵
Particle A is projected from the position 3, −1,2 with
velocity −1,6,2 .
At the same instant, particle B is projected from the position
5,5,4 with velocity −2,3,1 .
Do the particles collide?
𝐬𝐴 − 𝐬𝐵 = −2𝐢 − 6𝐣 − 2𝐤
𝐮𝐴 − 𝐮𝐵 = 1𝐢 + 3𝐣 + 1𝐤
The vector equation of a plane
𝐫. 𝐧 = 𝑑
1
Find the angle between the plane 𝐫. 4 = 2
−2
−4
2
and the line 𝐫 = 2 + λ 1
5
3
Dynamic lighting
Angle between
normal vector
and light
source
determines
how light the
surface should
appear
The dot product and computer games..
The dot product is the Swiss army knife of computer
games
It has hundreds of uses
Sometimes new ones are found accidentally
Angles: View Testing
Simple view culling
View vector 𝐯 and vector 𝐭 to object in scene
𝐭=𝐨−𝐞
If v • t < 0, object behind us, cull
e
v
t
o
Angles: Collision Response
Have normal 𝐧 (from object A to object B), relative
velocity 𝐯𝐴 − 𝐯𝐵
Three cases of contact:
– Separating
(𝐯𝐴 − 𝐯𝐵 ) • 𝐧 < 0
– Colliding
(𝐯𝐴 − 𝐯𝐵 ) • 𝐧 > 0
– Resting
(𝐯𝐴 − 𝐯𝐵) • 𝐧 = 0
𝐛
𝐎
𝐚
𝐛−𝐚
𝐛
𝐎
𝐚
𝐁
𝐛
𝐃
𝐛−𝐚
𝐆
𝐂
𝐅
𝐇
𝐎
𝐄
𝐀
𝐚
Area of OAB=
1
2
1
𝑂𝐶 = 𝐚 + 𝐛 − 𝐚 = 𝐚 + 𝐛
3
3
3
1
2
𝐴𝐷 = 𝐛 − 𝐚 − 𝐛 = −𝐚 + 𝐛
3
3
1
1
𝐵𝐸 = −𝐛 + 𝐚 = 𝐚 − 𝐛
3
3
1
2
𝐚×𝐛
2
1
𝑂𝐶 = 𝐚 + 𝐛
3
3
𝑙3
𝑙1 : 𝐫 = λ
2
𝐴𝐷 = −𝐚 + 𝐛
3
𝑙2
𝑙1
2
𝐚
3
1
𝐵𝐸 = 𝐚 − 𝐛
3
+
1
𝐛
3
2
3
𝑙2 : 𝐫 = 𝐚 + μ −𝐚 + 𝐛
𝑙3 : 𝐫 = 𝐛 + ν
1
𝐚
3
−𝐛
𝑙1 : 𝐫 = λ
𝑙3
𝑙2
𝑙1
Intersections
F: 𝑙1 and 𝑙2
2
𝐚
3
1
+ 𝐛
3
2
3
𝑙2 : 𝐫 = 𝐚 + μ −𝐚 + 𝐛
𝑙3 : 𝐫 = 𝐛 + ν
2
1
2
λ 𝐚 + 𝐛 = 𝐚 + μ −𝐚 + 𝐛
3
3
3
2
1
2
λ=1−μ
λ= μ
3
3
3
3
6
μ=
λ=
7
7
1
𝐚
3
−𝐛
𝑙1 : 𝐫 = λ
𝑙3
G: 𝑙2 and 𝑙3
1
+ 𝐛
3
2
3
𝑙2 : 𝐫 = 𝐚 + μ −𝐚 + 𝐛
𝑙2
𝑙1
Intersections
2
𝐚
3
𝑙3 : 𝐫 = 𝐛 + ν
2
𝐚 + μ −𝐚 + 𝐛
3
1
𝐚
3
1
=𝐛+ν 𝐚−𝐛
3
1
2
1−μ= ν
μ=1−ν
3
3
6
3
μ=
ν=
7
7
−𝐛
𝑙1 : 𝐫 = λ
𝑙3
𝑙2
𝑙1
Intersections
H: 𝑙1 and 𝑙3
2
𝐚
3
1
+ 𝐛
3
2
3
𝑙2 : 𝐫 = 𝐚 + μ −𝐚 + 𝐛
𝑙3 : 𝐫 = 𝐛 + ν
2
1
1
λ 𝐚+ 𝐛 =𝐛+ν 𝐚−𝐛
3
3
3
1
2
1
λ=1−ν
λ= ν
3
3
3
3
6
λ=
ν=
7
7
1
𝐚
3
−𝐛
4
2
𝑂𝐹 = 𝐚 + 𝐛
7
7
𝑙3
1
4
𝑂𝐺 = 𝐚 + 𝐛
7
7
2
1
𝑂𝐻 = 𝐚 + 𝐛
7
7
𝑙2
𝑙1
2
1
𝐻𝐹 = 𝐚 + 𝐛
7
7
1
3
𝐻𝐺 = − 𝐚 + 𝐛
7
7
1
Area 𝐻𝐹𝐺 =
2
2
1
1
3
𝐚+ 𝐛 × − 𝐚+ 𝐛
7
7
7
7
1
Area 𝐻𝐹𝐺 =
2
2
1
1
3
𝐚+ 𝐛 × − 𝐚+ 𝐛
7
7
7
7
1
− 𝐚
7
×
2
𝐚
7
1
+ 𝐛
7
−
3
+ 𝐛
7
𝟎
6
𝐚×𝐛
49
1
1
𝐛×𝐚 =
𝐚×𝐛
49
49
𝟎
1 7
Area 𝐻𝐹𝐺 =
𝐚×𝐛
2 49
1 1
=
𝐚×𝐛
7 2