VECTORS IN COMPONENT FORM
Vector as position vector of point A in 3 – D in Cartesian coordinate system:
𝑎𝑥
𝑎 = 𝑎 𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 ≡ 𝑎𝑦 ≡ 𝑎𝑥 𝑎𝑦 𝑎 𝑧
𝑎𝑧
where 𝑖, 𝑗 𝑎𝑛𝑑 𝑘 are unit vectors in x, y and z directions.
𝑖 =
1
0
0
𝑗 =
0
1
0
𝑘 =
0
0
1
Both, position vector of point A and point A have the same coordinates:
𝑎 =
𝑎 =
𝑎𝑥
𝑎𝑦 ,
𝑎𝑧
𝑎𝑥2 + 𝑎𝑦2 + 𝑎𝑧2
𝐴 = (𝑎𝑥 , 𝑎𝑦 , 𝑎𝑧 )
𝑎 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒, 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑟 𝑛𝑜𝑟𝑚
example:
2
2
0
0
𝑎= 7 = 0 + 7 + 0
−3
0
0
−3
1
0
0
= 2 0 + 7 1 − 3 0 = 2𝑖 + 7𝑗 − 3𝑘.
0
0
1
VECTOR BETWEEN TWO POINTS
𝑥𝐵 − 𝑥𝐴
𝐴𝐵 = 𝑦𝐵 − 𝑦𝐴 = (𝑥𝐵 − 𝑥𝐴 ) 𝑖 + (𝑦𝐵 − 𝑦𝐴 ) 𝑗 + (𝑧𝐵 − 𝑧𝐴 ) 𝑘
𝑧𝐵 − 𝑧𝐴
𝑥𝐴 − 𝑥𝐵
𝐵𝐴 = 𝑦𝐴 − 𝑦𝐵 = (𝑥𝐴 − 𝑥𝐵 ) 𝑖 + (𝑦𝐴 − 𝑦𝐵 ) 𝑗 + (𝑧𝐴 − 𝑧𝐵 ) 𝑘
𝑧𝐴 − 𝑧𝐵
𝑚𝑜𝑑𝑢𝑙𝑢𝑠 ≡ 𝑙𝑒𝑛𝑔𝑡ℎ:
=
𝐴𝐵 = 𝐵𝐴
(𝑥𝐵 − 𝑥𝐴 )2 +(𝑦𝐵 − 𝑦𝐴 )2 +(𝑧𝐵 − 𝑧𝐴 )2
Unit vector
Definition
It gives direction only!
A unit vector is a vector whose length is 1.
For a vector 𝑎 , a unit vector is in the
same direction as 𝑎 and is given by:
𝑎 =
𝑎 =
𝑎
𝑎
𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘
𝑎𝑥2 + 𝑎𝑦2 + 𝑎𝑧2
=
1
𝑎𝑥2 + 𝑎𝑦2 + 𝑎𝑧2
𝑎𝑥
𝑎𝑦
𝑎𝑧
𝑎𝑥 = (𝑥𝐵 − 𝑥𝐴 )
𝑎𝑦 = (𝑦𝐵 − 𝑦𝐴 )
𝑎𝑧 = (𝑧𝐵 − 𝑧𝐴 )
PARALLEL and COLLINEAR VECTORS
𝑎 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑏 ⇔ 𝑎 = 𝑘𝑏
𝑎=
6
9
3
𝑏=
2
3
1
6
9
3
2
= 3 3
1
𝑘𝜀𝑅
→ 𝑎 || 𝑏
𝑃𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑛𝑒𝑎𝑟 𝑖𝑓 𝑡ℎ𝑒𝑦 𝑙𝑖𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑙𝑖𝑛𝑒
𝐴, 𝐵 𝑎𝑛𝑑 𝐶 𝑎𝑟𝑒 𝑐𝑜𝑙𝑙𝑖𝑛𝑒𝑎𝑟 ⇔ 𝐴𝐵 = 𝑘𝐴𝐶 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑠𝑐𝑎𝑙𝑎𝑟 𝑘
(𝑜𝑛𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)
ARE 3 POINTS COLLINEAR ?
How can you check it:
1. Form two vectors with these three points.
They will definitely have one common point.
2. Check if these two vectors are parallel.
If two vectors have a common point and are parallel (or antiparallel)
∴ points are collinear.
Show that P(0, 2, 4), Q(10, 0, 0) and R(5, 1, 2) are collinear.
5
𝑃𝑅 = −1 ,
−2
𝑄𝑅 =
−5
1
2
𝑄𝑅 = −1 × 𝑃𝑅
𝑄𝑅 𝑎𝑛𝑑 𝑃𝑅 have a common direction and a common point.
Therefore P, Q and R are collinear.
THE DIVISION OF A LINE SEGMENT
X divides [AB]≡ AB in the ratio 𝑎: 𝑏 means 𝐴𝑋: 𝑋𝐵 = 𝑎 ∶ 𝑏
A = (2, 7, 8)
B = ( 2, 3, 12)
INTERNAL DIVISION
P divides [AB] internally
in ratio 1:3. Find P
𝐴𝑃: 𝑃𝐵 = 1: 3 → 𝐴𝑃 =
1
𝐴𝐵
4
𝐴𝑃 =
1
𝐴𝐵
4
𝑥−2
1 0
𝑦−7 =
−4
4
4
𝑧−8
point P is (2, 6, 9)
EXTERNAL DIVISION
X divide [AB] externally in ratio 2:1,
or
X divide [AB] in ratio –2:1. Find Q
𝐴𝑄: 𝑄𝐵 = −2: 1
𝐵𝑄 = 𝐴𝐵
𝑥−2
0
𝑦 − 3 = −4
4
𝑧 − 12
point Q is (2,– 1,16)
DOT/SCALAR PRODUCT
Scalar: ± 𝑛𝑢𝑚𝑏𝑒𝑟
Definition
The dot/scalar product of two vectors 𝑎 and 𝑏 is:
𝑎 • 𝑏 = 𝑏 • 𝑎 = 𝑎 𝑏 cos 𝜃
or: Product of the length of one of them and
projection of the other one on the first one
𝑏
𝑎
𝑏
𝑎
𝑏
θ
𝑎
In Cartesian coordinates:
𝑏𝑥
𝑎𝑥
𝑎 = 𝑎𝑦 and 𝑏 = 𝑏𝑦
𝑎𝑧
𝑏𝑧
𝑖 •𝑖 =1
𝑗•𝑗=1
𝑘•𝑘 =1 & 𝑖•𝑗=0
𝑖•𝑘 =0
𝑎 • 𝑏 = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 • 𝑏𝑥 𝑖 + 𝑏𝑦 𝑗 + 𝑏𝑧 𝑘 =
𝑎𝑥
→ 𝑎 • 𝑏 = 𝑎𝑦
𝑎𝑧
𝑏𝑥
𝑏𝑦
𝑏𝑧
= 𝑎𝑥 𝑏𝑥 + 𝑎𝑦 𝑏𝑦 + 𝑎𝑧 𝑏𝑧
𝑗•𝑘 =0
cos 900 = 0
Properties of dot product
∎ 𝑎 • 𝑏= 𝑏 • 𝑎
∎ 𝑖𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑡ℎ𝑒𝑛 𝑎 • 𝑏 = 𝑎 𝑏
∎ 𝑖𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑎𝑛𝑡𝑖𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑡ℎ𝑒𝑛 𝑎 • 𝑏 = − 𝑎 𝑏
∎ 𝑎 • 𝑎=
𝑎
2
∎ 𝑎 • 𝑏 + 𝑐 =𝑎•𝑏 + 𝑎•𝑐
∎
𝑎+𝑏
• 𝑐+𝑑 =𝑎•𝑐 + 𝑎•𝑑+𝑏•𝑐 + 𝑏•𝑑
∎ 𝑎 • 𝑏=0
𝑎 ≠ 0, 𝑏 ≠ 0 ↔ 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
CROSS / VECTOR PRODUCT
Definition
● The magnitude of the vector 𝑎 × 𝑏 is equal
to the area determined by both vectors.
𝑎 × 𝑏 = 𝑎 𝑏 𝑠𝑖𝑛 𝜃
● Direction of the vector 𝑎 × 𝑏 is given by right hand rule:
Point the fingers in direction of 𝑎; curl them toward 𝑏.
Your thumb points in the direction of cross product.
𝑏 × 𝑎 = −𝑎 × 𝑏
In Cartesian coordinates:
i×i=j× j=k ×k =0
i× j=k
j×k =i
k×i = j
𝑗
𝑖
𝑘
Using properties of determinates
𝑎
𝑐
𝑏
= 𝑎𝑑 − 𝑐𝑏
𝑑
We can write cross product in simple form:
+ −
𝑖
𝑎 × 𝑏 = 𝑎𝑥
𝑏𝑥
𝑗
𝑎𝑦
𝑏𝑦
𝑎𝑦
=𝑖 𝑏
𝑦
+
𝑘
𝑎𝑧
𝑏𝑧
𝑎𝑧
𝑎𝑥
𝑏𝑧 − 𝑗 𝑏𝑥
𝑎𝑥
𝑎𝑧
𝑏𝑧 + 𝑘 𝑏𝑥
𝑎𝑦
𝑏𝑦 =
𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦
𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧
𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥
Properties of vector/cross product
∎ 𝑎 × 𝑏 =−𝑏 × 𝑎
∎ 𝑖𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟, 𝑡ℎ𝑒𝑛 𝑎 × 𝑏 = 𝑎 𝑏
∎ 𝑎 × 𝑏 + 𝑐 =𝑎×𝑏 + 𝑎×𝑐
∎
𝑎+𝑏
× 𝑐+𝑑 = 𝑎×𝑐 + 𝑎×𝑑+𝑏×𝑐 + 𝑏×𝑑
∎ 𝑎 × 𝑏 = 0 𝑎 ≠ 0, 𝑏 ≠ 0 ↔ 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
𝐹𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑖𝑠 0.
𝑎 = 5𝑖 − 2𝑗 + 𝑘
𝑏 = 𝑖 + 𝑗 − 3𝑘
(a) Find the angle between them
(b) Find the unit vector perpendicular to both
(a) 𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛
𝑖
𝑎 ×𝑏 = 5
1
𝑎×𝑏
𝑎 𝑏
𝑗
−2
1
𝑘
1 = 5𝑖 + 16𝑗 + 7𝑘
−3
Find all vectors perpendicular to both
1
𝑎= 2
3
𝑖
𝑎×𝑏 = 1
3
3
𝑎𝑛𝑑 𝑏 = 2
1
−4
𝑗 𝑘
2 3 = 8
−4
2 1
⤇ 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑘 𝑖 − 2𝑗 + 𝑘
𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑛𝑜𝑛 − 𝑧𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟.
𝑎 × 𝑏 = 5𝑖 + 16𝑗 + 7𝑘 = 330
𝑎 = 30
𝑏 = 11
𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛 1 = 𝜋/2
(b) 𝑛 =
𝑎 ×𝑏
𝑎 ×𝑏
=
1
330
5
16
7
Find the area of the triangle with vertices A(1,1,3), B(4,-1,1), and C(0,1,8)
It is one-half the area of the parallelogram determined by the vectors
3
𝐴𝐵 = −2
−2
−1
and 𝐴𝐶 = 0
5
1
1
𝐴𝐵 × 𝐵𝐶 =
2
2
1
=
2
𝑖
3
−1
𝑗
−2
0
𝑘
−2
5
(−10)2 +(−13)2 +(−2)2
= 8.26 𝑢𝑛𝑖𝑡𝑠 2
1
⤇
2
−10
−13
−2
How do we use dot and cross product
• To find angle between vectors the easiest way is to use dot product, not vector product.
𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠
𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠
𝑏 • 𝑑
𝑏 𝑑
𝑏 • 𝑑
𝑏 𝑑
• Angle between vectors can
be acute or obtuse
• Angle between lines is by definition
acute angle between them, so
𝑏 𝑎𝑛𝑑 𝑑 are direction vectors
Dot product of perpendicular vectors is zero.
• To show that two lines are perpendicular use the dot product with line direction vectors.
• To show that two planes are perpendicular use the dot product on their normal vectors.
Volume of a parallelepiped = scalar triple product
𝑉 = 𝑐 ● 𝑎 ×𝑏
𝑐𝑥
= 𝑎𝑥
𝑏𝑥
𝑐𝑦
𝑎𝑦
𝑏𝑦
𝑐𝑧
𝑎𝑧 𝑢𝑛𝑖𝑡𝑠 3
𝑏𝑧
𝑐𝑥
𝑎𝑥
𝑏𝑥
𝑐𝑦
𝑎𝑦
𝑏𝑦
1
Volume of a tetrahedron = 6 scalar triple product
1
𝑉 = 𝑐 ● 𝑎 ×𝑏
6
1
=
6
𝑐𝑧
𝑎𝑧
𝑏𝑧
𝑢𝑛𝑖𝑡𝑠 3
TEST FOR FOUR COPLANAR POINTS
▪ 𝐹𝑜𝑢𝑟 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒 𝑐𝑜𝑝𝑙𝑎𝑛𝑎𝑟 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑒𝑡𝑟𝑎ℎ𝑒𝑑𝑟𝑜𝑛
𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒𝑚 𝑖𝑠 0:
Are the points A(1, 2, -4), B(3, 2, 0), C(2, 5, 1) and D(5, -3, -1) coplanar?
2
𝐴𝐵 = 0
4
1
𝐴𝐶 = 3
5
4
𝐴𝐷 = −5
3
⤇
2 0 4
𝐴𝐵 ● 𝐴𝐶 × 𝐴𝐷 = 1 3 5 = 2(9 + 25) + 4(−5 − 12) = 0
4 −5 3
∴
𝐴, 𝐵, 𝐶 𝑎𝑛𝑑 𝐷 𝑎𝑟𝑒 𝑐𝑜𝑝𝑙𝑎𝑛𝑎𝑟 𝑄. 𝐸. 𝐷
A line is completely determined by a fixed point and its direction.
Using vectors gives us a very neat way of writing down an equation which
gives the position vector of any point P on a given straight line. This method
works equally well in two or three dimensions.
LINE EQUATION IN 2 – D and 3 – D COORDINATE SYSTEM
● Vector equation of a line
The position vector 𝒓 of any general point P on the line passing through
point A and having direction vector 𝑏 is given by the equation
𝑟 =𝑎+𝑡𝑏
𝑡∈𝑅
IB Convention:
𝑟 = 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 + 𝜆 𝑏1 𝑖 + 𝑏2 𝑗 + 𝑏3 𝑘
𝑢𝑠𝑒 𝒕 𝑓𝑜𝑟 2 − 𝐷 𝑙𝑖𝑛𝑒
𝝀 𝑓𝑜𝑟 3 − 𝐷 𝑙𝑖𝑛𝑒
𝑎1
𝑏1
𝑥
𝑦 = 𝑎2 + 𝜆 𝑏2
𝑎3
𝑧
𝑏3
𝑜𝑟
● Parametric equation of a line – λ is called a parameter λ ∈ 𝑅
𝑎1
𝑏1
𝑥
𝑦 = 𝑎2 + 𝜆 𝑏2
𝑎3
𝑧
𝑏3
⇒
𝑥 = 𝑎1 + 𝜆𝑏1
𝑦 = 𝑎2 + 𝜆𝑏2
𝑧 = 𝑎3 + 𝜆𝑏3
● Cartesian equation of a line
𝑥 = 𝑎1 + 𝜆𝑏1 ⟹ 𝜆 = (𝑥 − 𝑎1 )/𝑏1
𝑦 = 𝑎2 + 𝜆𝑏2 ⟹ 𝜆 = (𝑦 − 𝑎2 )/𝑏2
𝑧 = 𝑎3 + 𝜆𝑏3 ⟹ 𝜆 = (𝑧 − 𝑎3 )/𝑏3
⟹
𝑥−𝑎1
𝑏1
=
𝑦−𝑎2
𝑏2
=
𝑧−𝑎3
𝑏3
(= 𝜆)
Find the equation of the line passing through the points A(3, 5, 2) and B(2, -4, 5).
Find the direction of the line:
One possible direction vector is
 2   3   1 
AB   4    5    9 
     
5   2 3 
The Cartesian equation of this line is
x3 y 5 z 2


1
9
3
(using the coordinates f point A).
The equivalent vector equation is
 x   3   1 
 y    5   t  9 
     
 z   2 3 
ANGLE BETWEEN TWO LINES
𝜃 = arccos
𝑏 • 𝑑
𝑏 𝑑
Two vectors
𝜃 = 𝑎𝑟𝑐 cos
𝑏 • 𝑑
𝑏 𝑑
Shortest distance from a point to a line
Point P is at the shortest distance from the line
when PQ is perpendicular to 𝑏
∴ 𝑃𝑄 • 𝑏 = 0
1
2
Find the shortest distance between 𝑟 = 3 + 𝜆 3
1
2
and point P (1,2,3).
(The goal is to find Q first, and then 𝑃𝑄 )
Point Q is on the line, hence its coordinates must satisfy line equation:
𝑥𝑄
1 + 2𝜆
2𝜆
𝑦𝑄 = 3 + 3𝜆 ⇒ 𝑃𝑄 = 1 + 3𝜆
𝑧𝑄
1 + 2𝜆
−2 + 2𝜆
2𝜆
2
⇒ 1 + 3𝜆 • 3 = 0
−2 + 2𝜆
2
⇒ 4𝜆 + 3 + 9𝜆 − 4 + 4𝜆 = 0
2/17
⇒ 𝑃𝑄 = 20/17
32/17
⇒ 𝑓𝑖𝑛𝑑 𝑃𝑄
⇒ 17 𝜆 = 1
⇒𝜆=
1
17
Relationship between lines
2 – D:
3 – D:
● the lines are coplanar (they lie in the same plane). They could be:
▪ intersecting
▪ parallel
▪ coincident
● the lines are not coplanar and are therefore skew
(neither parallel nor intersecting)
1
4
𝑟1 = 2 + λ 5
6
3
2
1
and 𝑟2 = − 2 + 𝜇 0
−1
3
.
Are the lines
∙ the same?…….check by inspection
∙ parallel?………check by inspection
∙ skew or do they have one point in common?
solving 𝑟1 = 𝑟2 will give 3 equations in  and µ.
Solve two of the equations for  and µ.
if the values of  and µ do not satisfy the third equation then
the lines are skew, and they do not intersect.
If these values do satisfy the three equations then substitute the value
of  or µ into the appropriate line and find the point of intersection.
Line 1: 𝑥 = −1 + 2𝑠,
Line 2: 𝑥 = 1 − 𝑡,
Line 3: 𝑥 = 1 + 2𝑢,
𝑦 = 1 − 2𝑠,
𝑦 = 𝑡,
𝑦 = −1 − 𝑢,
𝑧 = 1 + 4𝑠
𝑧 = 3 − 2𝑡
𝑧 = 4 + 3𝑢
a) Show that lines 2 and 3 intersect and find angle between them
b) Show that line 1 and 3 are skew.
a) 1 − 𝑡 = 1 + 2𝑢 ⇒ 𝑡 = −2𝑢,
𝑡 = −1 − 𝑢
⇒ −2𝑢 = −1 − 𝑢
⇒ 𝑢 = 1 & 𝑡 = −2
𝑐ℎ𝑒𝑐𝑘𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑧: 3 − 2𝑡 = 4 + 3𝑢 ⇒ 3 − 2 −2 = 4 + 3 1
𝑐𝑜𝑛𝑓𝑖𝑟𝑚𝑒𝑑 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 (3, −2, 7)
−1
𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑓𝑜𝑟 𝑙𝑖𝑛𝑒 2 𝑎𝑛𝑑 𝑙𝑖𝑛𝑒 3 𝑎𝑟𝑒: 𝑏 = 1
−2
cos 𝜃 =
𝑏•𝑑
𝑏 𝑑
=
−2−1−6
1+1+4 4+1+9
=
9
84
𝑏) − 1 + 2𝑠 = 1 + 2𝑢 ⇒ 2𝑠 − 2𝑢 = 2,
𝑐ℎ𝑒𝑐𝑘𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑧: 1 + 4𝑠 = 4 + 3𝑢
2
𝑎𝑛𝑑 𝑑 = −1
3
⇒ 𝜃 ≈ 10.9𝑜
1 − 2𝑠 = −1 − 𝑢
⇒ −2𝑠 + 𝑢 = −2,
⇒1+4 1 =4+3 0
⇒𝑢=0 & 𝑠=1
⇒5≠4
⇒ 𝑛𝑜 𝑠𝑖𝑚𝑢𝑙𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑡𝑜 𝑎𝑙𝑙 3 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
∴ 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑑𝑜 𝑛𝑜𝑡 𝑚𝑒𝑒𝑡, 𝑎𝑛𝑑 𝑎𝑠 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑏 ≠ 𝑘 𝑑 , 𝑘𝜖𝑅 𝑡ℎ𝑒𝑦 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑠𝑘𝑒𝑤.
Distance between two skew lines
𝑟 = 𝑎 + 𝜆 𝑏 𝑎𝑛𝑑 𝑟 = 𝑐 + 𝜇 𝑑
The cross product of 𝑏 and 𝑑 is perpendicular
to both lines, as is the unit vector:
𝑛=
𝑏×𝑑
𝑏×𝑑
The distance between the lines is then
𝑑 = 𝑛 • (𝑐 − 𝑎)
(sometimes I see it, sometimes I don’t)
PLANE EQUATION
● Vector equation of a plane 𝑟 = 𝑎 + 𝜆 𝑏 + µ𝑐
A plane is completely determined by two intersecting lines, what can
be translated into a fixed point A and two nonparallel direction vectors
The position vector 𝑟 of any general point P on the plane passing
through point A and having direction vectors 𝑏 and 𝑐 is given by the equation
𝑟 = 𝑎 + 𝜆 𝑏 + µ𝑐
𝜆, µ ∈ 𝑅
𝐴𝑃 = 𝜆 𝑏 + µ𝑐
● Parametric equation of a plane: λ , μ are called a parameters λ,μ ∈ 𝑅
𝑎1
𝑐1
𝑏1
𝑥
𝑦 = 𝑎2 + 𝜆 𝑏2 + 𝜇 𝑐2
𝑎3
𝑐3
𝑧
𝑏3
⇒
𝑥 = 𝑎1 + 𝜆𝑏1 + 𝜇𝑐1
𝑦 = 𝑎2 + 𝜆𝑏2 + 𝜇𝑐2
𝑧 = 𝑎3 + 𝜆𝑏3 + 𝜇𝑐3
● Normal/Scalar product form of vector equation of a plane
𝑛 • 𝑟 = 𝑛 • 𝑎 + 𝜆 𝑏 + µ𝑐
⇒
𝑟 • 𝑛 = 𝑎 • 𝑛 𝑜𝑟 𝑛 • 𝑟 − 𝑎 = 0
● Cartesian equation of a plane
𝑟•𝑛 =𝑎•𝑛
⤇
𝑛1 𝑥 + 𝑛2 𝑦 + 𝑛3 𝑧 = 𝑛1 𝑎1 + 𝑛2 𝑎2 + 𝑛3 𝑎3 = 𝑑
𝑛1 𝑥 + 𝑛2 𝑦 + 𝑛3 𝑧 = 𝑑
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑜𝑟𝑖𝑔𝑖𝑛:
𝐷=
=
𝑟•𝑛 = 𝑎•𝑛
𝑎•𝑛
𝑛12 +𝑛22 +𝑛32
=
𝑛1 𝑎1 +𝑛2 𝑎2 +𝑛3 𝑎3
𝑛12 +𝑛22 +𝑛32
What does the equation 3x + 4y = 12 give in 2 and 3 dimensions?
https://www.osc-ib.com/ib-videos/default.asp
http://www.globaljaya.net/secondary/IB/Subjects%20Report/May%202012%20
subject%20report/Maths%20HL%20subject%20report%202012%20TZ1.pdf
Find the equation of the plane passing through the
three points P1(1,-1,4), P2(2,7,-1), and P3(5,0,-1).
𝑏 = 𝑃1 𝑃2 =
1
8
−5
𝑐 = 𝑃1 𝑃3 =
1
𝑜𝑛𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑃1 = −1
4
4
1
−5
1
3
5
containing point (-2, 3, 4) .
1
3 • −2,3,4 = −2 + 9 + 20 = 27
5
𝑥 + 3𝑦 + 5𝑧 = 27
vector form:
1
𝑟 = −1
4
Find the equation of the plane with normal vector
+𝜆
1
4
8 +µ 1
−5
−5
−35
𝑖 𝑗 𝑘
𝑛 = 1 8 −5 = −15
−31
4 1 −5
3
Find the distance of the plane 𝑟● 2 = 8
−4
from the origin, and the unit vector
perpendicular to the plane.
3
2
−4
Any non-zero multiple of 𝑛 is also
a normal vector of the plane. Multiply by -1.
35
𝑛 = 15
31
1
35
15 • −1 = 144
4
31
𝐶𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑓𝑜𝑟𝑚:
35𝑥 + 15𝑦 + 31𝑧 = 144
= 29
3
𝑟● 2
29
−4
1
𝐷=
8
29
𝑛=
=
8
29
3
2
29 −4
1
ANGLES
● The angle between a line and a plane
𝑠𝑖𝑛 𝜃 = 𝑐𝑜𝑠 𝜙 =
𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛
𝑛•𝑑
𝑛 𝑑
𝑛•𝑑
𝑛 𝑑
● The angle between two planes
The angle between two planes is the same
as the angle between their 2 normal vectors
𝑐𝑜𝑠 𝜃 =
𝑛•𝑚
𝑛 𝑚
𝜃 = 𝑎𝑟𝑐 𝑐𝑜𝑠
𝑛•𝑚
𝑛 𝑚
take acute angle
● INTERSECTION OF TWO or MORE PLANES