VECTORS IN COMPONENT FORM - Uplift Education · PLANE EQUATION Vector equation of a plane N = + +µ...
Transcript of VECTORS IN COMPONENT FORM - Uplift Education · PLANE EQUATION Vector equation of a plane N = + +µ...
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VECTORS IN COMPONENT FORM
example:
𝑎 =27−3
=200
+070
+00−3
= 2100
+ 7010
− 3001
= 2 𝑖 + 7 𝑗 − 3 𝑘.
In Cartesian coordinates any 3 – D vector 𝑎 can be written as
𝑎 = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 ≡
𝑎𝑥
𝑎𝑦
𝑎𝑧
≡ 𝑎𝑥 𝑎𝑦 𝑎𝑧
where 𝑖, 𝑗 𝑎𝑛𝑑 𝑘 are unit vectors in x, y and z directions.
𝑖 =100
𝑗 =010
𝑘 =001
𝑎 = 𝑎𝑥2 + 𝑎𝑦
2 + 𝑎𝑧2 𝑎 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒, 𝑙𝑒𝑛𝑔𝑡ℎ,𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑟 𝑛𝑜𝑟𝑚
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Unit vector
For a vector 𝑎 , a unit vector is in the
same direction as 𝑎 and is given by:
𝑎 =𝑎
𝑎
A unit vector is a vector whose length is 1.
Definition
It gives direction only!
𝑎 =𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧
𝑘
𝑎𝑥2 + 𝑎𝑦
2 + 𝑎𝑧2
=1
𝑎𝑥2 + 𝑎𝑦
2 + 𝑎𝑧2
𝑎𝑥
𝑎𝑦
𝑎𝑧
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VECTOR BETWEEN TWO POINTS
𝐴𝐵 =
𝑥𝐵 − 𝑥𝐴
𝑦𝐵 − 𝑦𝐴
𝑧𝐵 − 𝑧𝐴= (𝑥𝐵 − 𝑥𝐴) 𝑖 + (𝑦𝐵 − 𝑦𝐴) 𝑗 + (𝑧𝐵 − 𝑧𝐴) 𝑘
𝐵𝐴 =
𝑥𝐴 − 𝑥𝐵
𝑦𝐴 − 𝑦𝐵
𝑧𝐴 − 𝑧𝐵= (𝑥𝐴 − 𝑥𝐵) 𝑖 + (𝑦𝐴 − 𝑦𝐵) 𝑗 + (𝑧𝐴 − 𝑧𝐵) 𝑘
𝑚𝑜𝑑𝑢𝑙𝑢𝑠 ≡ 𝑙𝑒𝑛𝑔𝑡ℎ: 𝐴𝐵 = 𝐵𝐴
= (𝑥𝐵 − 𝑥𝐴)2+(𝑦𝐵 − 𝑦𝐴)
2+(𝑧𝐵 − 𝑧𝐴)2
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PARALLEL VECTORS
𝑎 = 𝑘𝑏 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟. (𝑤ℎ𝑒𝑟𝑒 𝑘𝜀𝑅)
𝑎 =693
𝑏 =231
693
= 3231
→ 𝑎 || 𝑏
Two vectors 𝑎 and 𝑏 are parallel if and only if
Definition
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ARE 3 POINTS COLLINEAR ?
Show that P(0, 2, 4), Q(10, 0, 0) and R(5, 1, 2) are collinear.
𝑄𝑅 𝑎𝑛𝑑 𝑃𝑅 have a common direction and a common point.
Therefore P, Q and R are collinear.
𝑃𝑅 =5−1−2
, 𝑄𝑅 =−512
𝑄𝑅 = −1 × 𝑃𝑅
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)
∴ they are collinear.
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THE DIVISION OF A LINE SEGMENT
X divides [AB]≡ AB in the ratio 𝑎: 𝑏 means 𝐴𝑋:𝑋𝐵 = 𝑎 ∶ 𝑏
INTERNAL DIVISION
P divides [AB] internally
in ratio 1:3. Find P
EXTERNAL DIVISION
X divide [AB] externally in ratio 2:1,
or
X divide [AB] in ratio –2:1
A = (2, 7, 8) B = ( 2, 3, 12)
𝑥 − 2𝑦 − 7𝑧 − 8
+1
4
0−44𝐴𝑃: 𝑃𝐵 = 1: 3 → 𝐴𝑃 =
1
4𝐴𝐵
𝑥 − 2𝑦 − 3𝑧 − 12
=0−44
point P is (2, 6, 9)
point Q is (2,– 1,16)
𝐴𝑄:𝑄𝐵 = −2: 1
𝐴𝑃 =1
4𝐴𝐵
𝐵𝑄 = 𝐴𝐵
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DOT/SCALAR PRODUCT
The dot/scalar product of two vectors 𝑎 and 𝑏 is:
𝑎 • 𝑏 = 𝑏 • 𝑎 = 𝑎 𝑏 cos 𝜃
𝑎 • 𝑏 =
𝑎𝑥
𝑎𝑦
𝑎𝑧
𝑏𝑥
𝑏𝑦
𝑏𝑧
= 𝑎𝑥𝑏𝑥 + 𝑎𝑦𝑏𝑦 + 𝑎𝑧𝑏𝑧
or: Product of the length of one of them and
projection of the other one on the first one
Scalar: ± 𝑛𝑢𝑚𝑏𝑒𝑟
𝑎 𝑎
𝑏
𝑏
Definition
𝑎
𝑏
θ
Ex: Find the angle between
𝑎 = 5 𝑖 − 2 𝑗 + 𝑘 𝑎𝑛𝑑 𝑏 = 𝑖 + 𝑗 − 3 𝑘
𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠𝑎 • 𝑏
𝑎 𝑏=
0
30 11
𝜃 = 𝑎𝑟𝑐 𝑐𝑜𝑠 0 = 𝜋/2
Dot product of perpendicular
vectors is zero.
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𝑎 • 𝑏 =
𝑎𝑥
𝑎𝑦
𝑎𝑧
𝑏𝑥
𝑏𝑦
𝑏𝑧
= 𝑎𝑥𝑏𝑥 + 𝑎𝑦𝑏𝑦 + 𝑎𝑧𝑏𝑧
𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 • 𝑏𝑥 𝑖 + 𝑏𝑦 𝑗 + 𝑏𝑧
𝑘 =
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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
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The magnitude of the vector 𝑎 × 𝑏 is equal
to the area determined by both vectors.
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.
CROSS / VECTOR PRODUCT
𝑎 × 𝑏 = 𝑎 𝑏 𝑠𝑖𝑛 𝜃
𝑏 × 𝑎 = − 𝑎 × 𝑏
Definition
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∎ 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.
=>i × i = j × j = k × k = 0 i × j = k j × k = i k × i = j
⇒ 𝑎 × 𝑏 =
𝑎𝑦𝑏𝑧 − 𝑎𝑧𝑏𝑦
𝑎𝑧𝑏𝑥 − 𝑎𝑥𝑏𝑧
𝑎𝑥𝑏𝑦 − 𝑎𝑦𝑏𝑥
=
𝑖 𝑗 𝑘𝑎𝑥 𝑎𝑦 𝑎𝑧
𝑏𝑥 𝑏𝑦 𝑏𝑧
Properties of vector/cross product
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𝑎 = 5 𝑖 − 2 𝑗 + 𝑘
𝑏 = 𝑖 + 𝑗 − 3 𝑘
(a) Find the angle between them
(b) Find the unit vector perpendicular to both
(a) 𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛𝑎 × 𝑏
𝑎 𝑏
𝑎 × 𝑏 = 𝑖 𝑗 𝑘
5 −2 11 1 −3
= 5 𝑖 + 16 𝑗 + 7 𝑘
𝑎 × 𝑏 = 5 𝑖 + 16 𝑗 + 7 𝑘 = 330
𝑎 = 30 𝑏 = 11
𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛 1 = 𝜋/2
(b) 𝑛 =𝑎 ×𝑏
𝑎 ×𝑏=
1
330
5167
Find all vectors perpendicular to both
𝑎 =123
𝑎𝑛𝑑 𝑏 =321
𝑎 × 𝑏 = 𝑖 𝑗 𝑘
1 2 33 2 1
=−48−4
⤇ 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑘 𝑖 − 2 𝑗 + 𝑘
𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑛𝑜𝑛 − 𝑧𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟.
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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
𝐴𝐵 =3−2−2
and 𝐴𝐶 =−105
1
2𝐴𝐵 × 𝐵𝐶 =
1
2
𝑖 𝑗 𝑘3 −2 −2−1 0 5
⤇1
2
−10−13−2
=1
2(−10)2+(−13)2+(−2)2
= 8.26 𝑢𝑛𝑖𝑡𝑠2
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• To find angle between vectors the easiest way is to use dot product,
not vector product.
• Angle between vector can be positive or negative
• Angle between lines is by definition acute angle between them, so
Dot product of perpendicular vectors is zero.
For perpendicular vectors the dot/scalar product is 0.
𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠 𝑎 • 𝑏
𝑎 𝑏
𝜃 = 𝑎𝑟𝑐𝑐𝑜𝑠 𝑎 • 𝑏
𝑎 𝑏
Conclusions:
• To show that two lines are perpendicular usethe dot product with line direction vectors.
• To show that two planes are perpendicularuse the dot product on their normal vectors.
• To find the angle between two lines
cos 𝜃 =𝑎 • 𝑏
𝑎 𝑏
𝑎 𝑎𝑛𝑑 𝑏 are direction vectors
acute angle !
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Volume of a parallelepiped = scalar triple product
𝑉 = 𝑐 ● 𝑎 × 𝑏 =
𝑐𝑥 𝑐𝑦 𝑐𝑧𝑎𝑥 𝑎𝑦 𝑎𝑧
𝑏𝑥 𝑏𝑦 𝑏𝑧
𝑢𝑛𝑖𝑡𝑠3
Volume of a tetrahedron = 1
6scalar triple product
𝑉 =1
6 𝑐 ● 𝑎 × 𝑏 =
1
6
𝑐𝑥 𝑐𝑦 𝑐𝑧𝑎𝑥 𝑎𝑦 𝑎𝑧
𝑏𝑥 𝑏𝑦 𝑏𝑧
𝑢𝑛𝑖𝑡𝑠3
TEST FOR FOUR COPLANAR POINTS
If the volume of the tetrahedron is zero points are coplanar.
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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 + 𝑡 𝑏 𝑡 ∈ 𝑅IB Convention: 𝑢𝑠𝑒 𝒕 𝑓𝑜𝑟 2 − 𝐷 𝑙𝑖𝑛𝑒
𝝀 𝑓𝑜𝑟 3 − 𝐷 𝑙𝑖𝑛𝑒
𝑟 = 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 + 𝜆 𝑏1 𝑖 + 𝑏2 𝑗 + 𝑏3
𝑘 𝑜𝑟𝑥𝑦𝑧
=
𝑎1
𝑎2
𝑎3
+ 𝜆
𝑏1
𝑏2
𝑏3
● Parametric equation of a line – λ is called a parameter λ ∈ 𝑅
𝑥𝑦𝑧
=
𝑎1
𝑎2
𝑎3
+ 𝜆
𝑏1
𝑏2
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)
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Find a) vector b) parametric c) Cartesian and d) general line
equation of a line passing through 𝐴 =12
& 𝐵 =36
.
choosing point 𝐴 =12
and direction 𝐴𝐵 =24
𝑎)𝑥𝑦 =
12
+ 𝑡24
𝑡 ∈ 𝑅
𝑏)𝑥 = 1 + 2𝑡𝑦 = 2 + 4𝑡
𝑡 ∈ 𝑅
𝑐)𝑥−1
2=
𝑦−2
4
𝑑) 4𝑥 − 4 = 2𝑦 − 4 ⟹ 2𝑦 − 4𝑥 = 0
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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
Find the shortest distance between 𝑟 =131
+ 𝜆232
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
𝑦𝑄
𝑧𝑄=
1 + 2𝜆3 + 3𝜆1 + 2𝜆
⇒ 𝑃𝑄 =2𝜆
1 + 3𝜆−2 + 2𝜆
⇒2𝜆
1 + 3𝜆−2 + 2𝜆
•232
= 0
⇒ 4𝜆 + 3 + 9𝜆 − 4 + 4𝜆 = 0 ⇒ 17 𝜆 = 1 ⇒ 𝜆 =1
17
⇒ 𝑃𝑄 =
2/1720/1732/17
⇒ 𝑓𝑖𝑛𝑑 𝑃𝑄
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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)
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𝑟1 =123
+ λ456
and 𝑟2 = −12−1
+ 𝜇203
.
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.
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Line 1: 𝑥 = −1 + 2𝑠, 𝑦 = 1 − 2𝑠, 𝑧 = 1 + 4𝑠Line 2: 𝑥 = 1 − 𝑡, 𝑦 = 𝑡, 𝑧 = 3 − 2𝑡Line 3: 𝑥 = 1 + 2𝑢, 𝑦 = −1 − 𝑢, 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)
𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑓𝑜𝑟 𝑙𝑖𝑛𝑒 2 𝑎𝑛𝑑 𝑙𝑖𝑛𝑒 3 𝑎𝑟𝑒: 𝑏 =−11−2
𝑎𝑛𝑑 𝑑 =2−13
cos 𝜃 =𝑏 • 𝑑
𝑏 𝑑=
−2−1−6
1+1+4 4+1+9=
9
84⇒ 𝜃 ≈ 10.9𝑜
𝑏) − 1 + 2𝑠 = 1 + 2𝑢 ⇒ 2𝑠 − 2𝑢 = 2, 1 − 2𝑠 = −1 − 𝑢 ⇒ −2𝑠 + 𝑢 = −2, ⇒ 𝑢 = 0 & 𝑠 = 1
𝑐ℎ𝑒𝑐𝑘𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑧: 1 + 4𝑠 = 4 + 3𝑢 ⇒ 1 + 4 1 = 4 + 3 0 ⇒ 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.
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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)
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PLANE EQUATION
● Vector equation of a plane 𝑟 = 𝑎 + 𝜆 𝑏 + µ 𝑐
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 + µ 𝑐 𝜆, µ ∈ 𝑅 𝐴𝑃 = 𝜆 𝑏 + µ 𝑐
● Parametric equation of a plane: λ , μ are called a parameters λ,μ ∈ 𝑅
𝑥𝑦𝑧
=
𝑎1
𝑎2
𝑎3
+ 𝜆
𝑏1
𝑏2
𝑏3
+ 𝜇
𝑐1𝑐2𝑐3
⇒
𝑥 = 𝑎1 + 𝜆𝑏1 + 𝜇𝑐1𝑦 = 𝑎2 + 𝜆𝑏2 + 𝜇𝑐2𝑧 = 𝑎3 + 𝜆𝑏3 + 𝜇𝑐3
● Normal/Scalar product form of vector equation of a plane
𝑛 • 𝑟 = 𝑛 • 𝑎 + 𝜆 𝑏 + µ 𝑐 ⇒ 𝑟 • 𝑛 = 𝑎 • 𝑛 𝑜𝑟 𝑛 • 𝑟 − 𝑎 = 0
● Cartesian equation of a plane
𝑟 • 𝑛 = 𝑎 • 𝑛 ⤇ 𝑛1𝑥 + 𝑛2𝑦 + 𝑛3𝑧 = 𝑛1𝑎1 + 𝑛2𝑎2 + 𝑛3𝑎3 = 𝑑
𝑛1𝑥 + 𝑛2𝑦 + 𝑛3𝑧 = 𝑑
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𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑜𝑟𝑖𝑔𝑖𝑛:
𝐷 = 𝑟 • 𝑛 = 𝑎 • 𝑛
=𝑎•𝑛
𝑛12+𝑛2
2+𝑛32
=𝑛1𝑎1+𝑛2𝑎2+𝑛3𝑎3
𝑛12+𝑛2
2+𝑛32
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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 = 𝑃1𝑃2 =18−5
𝑐 = 𝑃1𝑃3 =41−5
𝑜𝑛𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑃1 =1−14
vector form:
𝑛 = 𝑖 𝑗 𝑘
1 8 −54 1 −5
=−35−15−31
Any non-zero multiple of 𝑛 is also
a normal vector of the plane. Multiply by -1.
𝑛 =351531
𝑟 =1−14
+ 𝜆18−5
+ µ41−5
351531
•1−14
= 144
𝐶𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑓𝑜𝑟𝑚:35𝑥 + 15𝑦 + 31𝑧 = 144
Find the equation of the plane with normal vector
135
containing point (-2, 3, 4) .
135
• −2,3,4 = −2 + 9 + 20 = 27
𝑥 + 3𝑦 + 5𝑧 = 27
Find the distance of the plane 𝑟●32−4
= 8
from the origin, and the unit vector
perpendicular to the plane.
32−4
= 29
1
29 𝑟●
32−4
=8
29
𝐷 =8
29 𝑛 =
1
29
32−4
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It's fairly straightforward to convert a vector equation into a Cartesian equation, as you simply find the
cross product of the two vectors appearing in the vector equation to find a normal to the plane and use
that to find the Cartesian equation. But this process can't exactly be reversed to go the other way.
∙ To convert Cartesian -> vector form, you need either two vectors or three points that lie on the plane!
Convert the Cartesian equation of the plane x -2y + 2z = 5 into
(a) a vector equation of the form 𝑟 • 𝑛 = 𝐷, where 𝑛 is a unit vector.
(b) vector form
(c) State the perpendicular distance of the plane from the origin.
(a) 𝑛 =1
1+4+4
1−22
=1
3
1−22
⟹𝑥𝑦𝑧
●
1/3−2/32/3
= 5/3
5/3 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛
(b) choose three arbitrary random non-collinear points: 𝐴(0, 1/2, 3) 𝐵(0, 3/2, 4) 𝐶(3, 0, 1)
𝐴𝐵 =011
𝐴𝐶 =3
−1/2−2
⟹ 𝑟 =0
1/23
+ 𝜆011
+ 𝜇3
−1/2−2
(c) 5/3
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ANGLES
● The angle between a line and a plane
𝑠𝑖𝑛 𝜃 = 𝑐𝑜𝑠 𝜙 =𝑛 • 𝑑
𝑛 𝑑
take acute angle
𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛𝑛• 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 𝑐𝑜𝑠𝑛 • 𝑚
𝑛 𝑚
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INTERSECTION of a LINE and a PLANE
First check that the line is not contained in the plane, nor parallel to it.
i.e456
•123
= 32
Substitute the line equation into the plane equation to obtain the value of the
line parameter, µ. Substitute for µ into the equation of the line to obtain the
co-ordinates of the point of intersection.
i.e. Solve
1 + 4𝜇−2 + 5𝜇−1 + 6𝜇
123
= 5 1 + 4µ - 4+10µ -3 + 18µ = 5.
Solve for µ and substitute into the equation of the line to get the point of intersection.
If this equation gives you something like 0 = 5, then the line will be parallel and not in the plane,
and if the equation gives you something like 5 = 5 then the line is contained in the plane.
Line L: 𝑟 =1−2−1
+ 𝜇456
and plane Π: 𝑥 + 2𝑦 + 3𝑧 = 5
therefore the line and the plane are not parallel and the line will intersect the plane in one point.
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INTERSECTION of TWO PLANES and the EQUATION of the LINE of INTERSECTION
Solve between the two plane equations in terms of a parameter say, λ,
plane Π1: 𝑥 + 2𝑦 + 3𝑧 = 5 and plane Π2: 2𝑥 − 2𝑦 − 2𝑧 = 2
First, check by inspection, that the planes are not parallel (normal vectors are not parallel).
Find intersection:
0 0 01 2 32 −2 −2
052
~ 𝑅2 = 𝑅2 +3
2𝑅3 ~
0 0 04 −1 02 −2 −2
082
⤇ 𝑥 = 𝑡 𝑦 = 4𝑥 + 8 𝑧 = −3𝑥 + 7
⤇ 𝑡 = 𝑥 =𝑦 + 8
4=
𝑧 − 7
−3
which is the equation of the common line, which in vector form is 𝑟 =0−87
+ 𝑡14−3
(Equally you can write t = y as a function of x and z, or t = z as a function of x and y).
OR Find the vector product of both normals to give the direction of the line.
Then you need a point on the line – do it in the future
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● INTERSECTION OF TWO or MORE PLANES
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What does the equation 3x + 4y = 12 give in 2 and 3 dimensions?
http://www.globaljaya.net/secondary/IB/Subjects%20Report/May%202012%20subject%20report/Maths%20HL%20subject%20report%202012%20TZ1.pdf
https://www.osc-ib.com/ib-videos/default.asp