1. 3d Vectors

2. Vector Equation of line
2d Vectors
Topic
3d Vectors
𝑎 1 i + 𝑎 2 j
Component form
𝑎 1 i + 𝑎 2 j + 𝑎 3 k
𝑎 1 𝑎 2
Column Vector
𝑎 1 𝑎 2 𝑎 3
𝒂 = 𝑎 𝑎 2 2
Length (magnitude)
𝒂 = 𝑎 𝑎 𝑎 3 2
Angle between vectors
a = 𝑎 1 i + 𝑎 2 j + 𝑎 3 k
b = 𝑏 1 i + 𝑏 2 j + 𝑏 3 k
𝑎 1 𝑏 1 + 𝑎 2 𝑏 2
Scalar product (a.b)
𝑎 1 𝑏 1 + 𝑎 2 𝑏 2 + 𝑎 3 𝑏 3
𝑟= 𝑎 1 𝑎 2 +𝜇 𝑢 1 𝑢 2
Vector Equation of line
𝑟= 𝑎 1 𝑎 2 𝑎 3 +𝜇 𝑢 1 𝑢 2 𝑢 3
𝑥− 𝑎 1 𝑢 1 = 𝑦− 𝑎 2 𝑢 2
Cartesian Equation
𝑥− 𝑎 1 𝑢 1 = 𝑦− 𝑎 2 𝑢 2 = 𝑧− 𝑎 3 𝑢 3
𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 𝒃
𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 𝒃

4. Answer for Q6

6. 𝑟= 𝑂𝐴 + 𝐴𝐵 +µ 𝐴𝐶 Vector Equation of a plane 𝑟=𝒂+ 𝒃−𝒂 + 𝜇(𝒄−𝒂)
The equation of a plane needs to tell you a start point on the plane and then the direction it travels in 2 dimensions.
𝑟= 𝑂𝐴 𝐴𝐵 +µ 𝐴𝐶 x y z
𝑟=𝒂+ 𝒃−𝒂 + 𝜇(𝒄−𝒂) A B C

7. Vector Equation of a plane 𝑟=𝒂+ 𝒃−𝒂 + 𝜇(𝒄−𝒂)
𝑟=𝒂+ 𝒃−𝒂 + 𝜇(𝒄−𝒂)
Find the vector equation of the plane through the points
a = (5, 3, 1) b = (0, 3, -2) c = (2, -4, 2)
𝑟= −5 0 −3 + 𝜇 −3 −7 1

8. Cartesian form for Equation of a plane
𝑟= −5 0 −3 + 𝜇 −3 −7 1
Cartesian form for Equation of a plane
Find the vector equation of the plane through the points
a = (5, 3, 1) b = (0, 3, -2) c = (2, -4, 2)
𝜇= 𝑦 −3 −7
𝑥 𝑦 𝑧 = −5 0 −3 + 𝜇 −3 −7 1
If you rearrange 2nd and substitute into the 3rd equation you get that:
𝑥 −5=−5 −3𝜇
= −7𝑧 −𝑦+10 21
𝑦 −3=−7𝜇
Substituting this and your rearranged 2nd equation into the 1st equation allows you to get an equation only in x, y and z
𝑧 −1=−3 +1𝜇
−21𝑥+14𝑦+35𝑧+28=0

9. Perpendicular to the plane (normal)
𝒏= 𝑛 1 𝑛 2 𝑛 3 n R A
𝑛 1 .𝑥+ 𝑛 2 .𝑦+ 𝑛 3 .𝑧+𝑑=0
Where the constant is defined by d = - a.n
Write down the equation of the plane through the point (5, -4, 6) given that the vector
is perpendicular to the plane
−3 2 4
𝑛 1 .𝑥+ 𝑛 2 .𝑦+ 𝑛 3 .𝑧
d = - a.n
−3 ×5+2×−4+4 ×6=1
−3𝑥+2𝑦+4𝑧
−𝟑𝒙+𝟐𝒚+𝟒𝒛−𝟏=𝟎

10. Intersection of a line and plane
Find the intersection of the line below with the plane –3x + 2y + 4z + 1 = 0.
−3 5−5𝜇 −3𝜇 +1=0
𝑟= 𝜇 −5 0 −3
−15+15𝜇+6+4 −12𝜇+1=0
3𝜇=4
𝜇= 4 3
𝑥 𝑦 𝑧 = 𝜇 −5 0 −3
Substitute into vector equation of line
𝑟= −5 0 −3 = −5/3 3 −3
𝑥=5 −5𝜇
Check in equation of plane
𝑦=3
−3 −5/ −3 +1=0
𝑧=1 −3𝜇

11. Distance of a point from a plane
The shortest distance of n from the plane will always be the perpendicular length to point A. n R A

12. Worked Example
A is the point (7, 5, 3) and the plane π has the equation 3x + 2y + z = 6. Find:
The equation of the line through A perpendicular to the plane
The point of intersection, P, of this line with the plane
The distance AP
a) The direction perpendicular to the plane 3x + 2y + z = 6 is
So the line through (7, 5, 3) perpendicular to the plane is given by:
𝑟= 𝜇 n R A
𝒏= 𝑛 1 𝑛 2 𝑛 3
𝑛 1 .𝑥+ 𝑛 2 .𝑦+ 𝑛 3 .𝑧+𝑑=0

13. Worked Example
A is the point (7, 5, 3) and the plane π has the equation 3x + 2y + z = 6. Find:
The equation of the line through A perpendicular to the plane
The point of intersection, P, of this line with the plane
The distance AP
b) For any point on the line substitute in:
3 7+3𝜇 𝜇 + 3+𝜇 =6
𝑥=7+3𝜇
𝑦=5+2𝜇
𝜇=−2
𝑧=3+ 𝜇
𝑟= −
𝑝=(1, 1, 1)

14. Worked Example
A is the point (7, 5, 3) and the plane π has the equation 3x + 2y + z = 6. Find:
The equation of the line through A perpendicular to the plane
The point of intersection, P, of this line with the plane
The distance AP
c) The vector 𝐴𝑃 is given by:
− = −6 −4 −2
−6 2 + −4 2 + −2 2 = 56
Please note that there is a generalised formula for distance of a point from a plane

15. 21+10+3−6 9+4+1 = 28 14 = 56 Worked Example 𝛼 𝛽 𝛾
A is the point (7, 5, 3) and the plane π has the equation 3x + 2y + z = 6.
𝛼 𝛽 𝛾
𝑛 1 .𝑥+ 𝑛 2 .𝑦+ 𝑛 3 .𝑧+𝑑=0
Please note that there is a generalised formula for distance of a point from a plane
𝛼𝑛 1 + 𝛽𝑛 2 + 𝛾𝑛 3 +𝑑 𝑛 𝑛 𝑛 3 2
− = = 56

16. Vector Equation of line
2d Vectors
Topic
3d Vectors
𝑎 1 i + 𝑎 2 j
Component form
𝑎 1 i + 𝑎 2 j + 𝑎 3 k
𝑎 1 𝑎 2
Column Vector
𝑎 1 𝑎 2 𝑎 3
𝒂 = 𝑎 𝑎 2 2
Length (magnitude)
𝒂 = 𝑎 𝑎 𝑎 3 2
Angle between vectors
a = 𝑎 1 i + 𝑎 2 j + 𝑎 3 k
b = 𝑏 1 i + 𝑏 2 j + 𝑏 3 k
𝑎 1 𝑏 1 + 𝑎 2 𝑏 2
Scalar product (a.b)
𝑎 1 𝑏 1 + 𝑎 2 𝑏 2 + 𝑎 3 𝑏 3
𝑟= 𝑎 1 𝑎 2 +𝜇 𝑢 1 𝑢 2
Vector Equation of line
𝑟= 𝑎 1 𝑎 2 𝑎 3 +𝜇 𝑢 1 𝑢 2 𝑢 3
𝑥− 𝑎 1 𝑢 1 = 𝑦− 𝑎 2 𝑢 2
Cartesian Equation
𝑥− 𝑎 1 𝑢 1 = 𝑦− 𝑎 2 𝑢 2 = 𝑧− 𝑎 3 𝑢 3
Vector Equation of a plane
𝑟=𝒂+ 𝒃−𝒂 + 𝜇(𝒄−𝒂)
Distance of a point from a plane
𝛼 𝛽 𝛾
𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 𝒃
𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 𝒃
𝛼𝑛 1 + 𝛽𝑛 2 + 𝛾𝑛 3 +𝑑 𝑛 𝑛 𝑛 3 2
Perpendicular to the plane (normal) n R A
𝒏= 𝑛 1 𝑛 2 𝑛 3
𝑛 1 .𝑥+ 𝑛 2 .𝑦+ 𝑛 3 .𝑧+𝑑=0
Where the constant is defined by d = - a.n