where ) - Uplift Education...In Cartesian coordinates 3 – D vector can be written as ⃗ = ̂+ ̂+...
Transcript of where ) - Uplift Education...In Cartesian coordinates 3 – D vector can be written as ⃗ = ̂+ ̂+...
1
In Cartesian coordinates 3 – D vector �⃗� can be written as
�⃗� = 𝑎𝑥𝑖̂ + 𝑎𝑦𝑗̂ + 𝑎𝑧�̂� ≡ (
𝑎𝑥
𝑎𝑦
𝑎𝑧
) ≡ (𝑎𝑥 𝑎𝑦 𝑎𝑧)
where 𝑖,̂ 𝑗̂ 𝑎𝑛𝑑 �̂� are unit vectors in x, y and z directions.
𝑖̂ = (100
) 𝑗̂ = (010
) �̂� = (001
)
|�⃗�| = √𝑎𝑥2 + 𝑎𝑦
2 + 𝑎𝑧2
|�⃗�| 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒, 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑟 𝑛𝑜𝑟𝑚
A unit vector is a vector whose length is 1. It gives direction only!
For a vector �⃗� , a unit vector is in the same direction as �⃗� given by:
�̂� = �⃗⃗�
|�⃗⃗�|
Two vectors �⃗� and �⃗⃗� are parallel if and only if
�⃗� = 𝑘�⃗⃗� 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟.
𝑤ℎ𝑒𝑟𝑒 𝑘𝜀𝑅
Show that P(0, 2, 4), Q(10, 0, 0) and R(5, 1, 2) are collinear.
𝑄𝑅⃗⃗ ⃗⃗ ⃗⃗ 𝑎𝑛𝑑 𝑃𝑅⃗⃗⃗⃗⃗⃗ have a common direction and a common point. ∴ Therefore P, Q and R are collinear.
● VECTORS IN COMPONENT FORM
● Unit vector
● VECTOR BETWEEN TWO POINTS
● PARALLEL VECTORS
● COLINEAR 3 POINTS
𝐴𝐵⃗⃗⃗⃗ ⃗⃗ = (
𝑥𝐵 − 𝑥𝐴
𝑦𝐵 − 𝑦𝐴
𝑧𝐵 − 𝑧𝐴
) = (𝑥𝐵 − 𝑥𝐴) 𝑖 ̂ + (𝑦𝐵 − 𝑦𝐴) 𝑗̂ + (𝑧𝐵 − 𝑧𝐴) �̂�
𝐵𝐴⃗⃗⃗⃗ ⃗⃗ = (
𝑥𝐴 − 𝑥𝐵
𝑦𝐴 − 𝑦𝐵
𝑧𝐴 − 𝑧𝐵
) = (𝑥𝐴 − 𝑥𝐵) 𝑖 ̂ + (𝑦𝐴 − 𝑦𝐵) 𝑗̂ + (𝑧𝐴 − 𝑧𝐵) �̂�
𝑚𝑜𝑑𝑢𝑙𝑢𝑠 ≡ 𝑙𝑒𝑛𝑔𝑡ℎ: | 𝐴𝐵⃗⃗⃗⃗ ⃗⃗ | = |𝐵𝐴⃗⃗⃗⃗ ⃗⃗ |
= √(𝑥𝐵 − 𝑥𝐴)2 + (𝑦𝐵 − 𝑦𝐴)2 + (𝑧𝐵 − 𝑧𝐴)2
Any vector can be written in
terms of 𝑖,̂ 𝑗̂ 𝑎𝑛𝑑 �̂�.
example:
(27
−3) = (
200
) + (070
) + (00
−3)
= 2 (100
) + 7 (010
) + 8 (001
)
= 2𝑖̂ + 7𝑗̂ + 8�̂�.
�⃗� = (693
) �⃗⃗� = (231
)
(693
) = 3 (231
) → �⃗� || �⃗⃗�
𝑃𝑅⃗⃗⃗⃗⃗⃗ = (5
−1−2
), 𝑄𝑅⃗⃗ ⃗⃗ ⃗⃗ = (−512
)
𝑃𝑅⃗⃗⃗⃗⃗⃗ = −1 × 𝑄𝑅⃗⃗⃗⃗ ⃗⃗
2
X divides [AB] in the ratio 𝑎: 𝑏 means 𝐴𝑋⃗⃗⃗⃗ ⃗⃗ : 𝑋𝐵⃗⃗ ⃗⃗ ⃗⃗ = 𝑎 ∶ 𝑏
INTERNAL DIVISION
We say X divide [AB] internally in ratio 1.5 : 1
ETERNAL DIVISION
We say X divide [AB] externally in ratio 2:1, or We say X divide [AB] in ratio –2:1 or 2: –1
I. Multiplying vector �⃗� by a scalar k:
𝑘�⃗� is a vector in the same direction as �⃗� with the magnitude stretched by a factor of k.
2. Opposite vector −�⃗� = (−1)�⃗� has direction opposite to �⃗� .
3. Addition graphically (head to tail method ): �⃗� + �⃗⃗�
4. subtraction graphically: �⃗� − �⃗⃗� = �⃗� + (−�⃗⃗�)
● THE DIVISION OF A LINE SEGMENT
● OPERATIONS ON VECTORS
If A is (2, 7, 8) and B is ( 2, 3, 12) find:
a. P if P divides [AB] in the ratio 1:3
b. Q if Q divides [AB] externally in the ratio 2:1.
𝑎. 𝐴𝑃⃗⃗⃗⃗⃗⃗ : 𝑃𝐵⃗⃗⃗⃗⃗⃗ = 1: 3
𝐴𝑃⃗⃗⃗⃗⃗⃗ =1
4 𝐴𝐵⃗⃗⃗⃗ ⃗⃗
(𝑥 − 2𝑦 − 7𝑧 − 8
) =1
4(
0−44
)
point P is (2, 6, 9)
𝑏. 𝐴𝑄⃗⃗ ⃗⃗ ⃗⃗ : 𝑄𝐵⃗⃗ ⃗⃗ ⃗⃗ = −2: 1
𝐵𝑄⃗⃗ ⃗⃗ ⃗⃗ = 𝐴𝐵⃗⃗⃗⃗ ⃗⃗
(𝑥 − 2𝑦 − 3
𝑧 − 12) = (
0−44
)
point Q is (2, – 1, 16)
3 5. Addition or subtraction analytically
𝐶 = 𝐴 + �⃗⃗� 𝐶𝑥 = 𝐴𝑥 + 𝐵𝑥 = 𝐴 cos 𝜃𝐴 + 𝐵𝑐𝑜𝑠 𝜃𝐵
𝐶𝑦 = 𝐴𝑦 + 𝐵𝑦 = 𝐴 𝑠𝑖𝑛 𝜃𝐴 + 𝐵𝑠𝑖𝑛 𝜃𝐵
𝐶 = √𝐶𝑥2 + 𝐶𝑦
2 ; from the picture
The dot/scalar product of two vectors �⃗� = (
𝑎𝑥
𝑎𝑦
𝑎𝑧
) and �⃗⃗� = (
𝑏𝑥
𝑏𝑦
𝑏𝑧
) is a scalar quantity (a real number)
�⃗� • �⃗⃗� = �⃗⃗� • �⃗� = |�⃗�||�⃗⃗�| cos 𝜃
�⃗� • �⃗⃗� = (
𝑎𝑥
𝑎𝑦
𝑎𝑧
) (
𝑏𝑥
𝑏𝑦
𝑏𝑧
) = (𝑎𝑥 𝑎𝑦 𝑎𝑧) (
𝑏𝑥
𝑏𝑦
𝑏𝑧
) = 𝑎𝑥𝑏𝑥 + 𝑎𝑦𝑏𝑦 + 𝑎𝑧𝑏𝑧
Properties of dot product
∎ �⃗� • �⃗⃗� = �⃗⃗� • �⃗�
∎ 𝑖𝑓 �⃗� 𝑎𝑛𝑑 �⃗⃗� 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑡ℎ𝑒𝑛 �⃗� • �⃗⃗� = |�⃗�||�⃗⃗�|
∎ 𝑖𝑓 �⃗� 𝑎𝑛𝑑 �⃗⃗� 𝑎𝑟𝑒 𝑎𝑛𝑡𝑖𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑡ℎ𝑒𝑛 �⃗� • �⃗⃗� = − |�⃗�||�⃗⃗�|
∎ �⃗� • �⃗� = |�⃗� |2
∎ �⃗� • (�⃗⃗� + 𝑐) = �⃗� • �⃗⃗� + �⃗� • 𝑐
∎ (�⃗� + �⃗⃗�) • (𝑐 + 𝑑) = �⃗� • 𝑐 + �⃗� • 𝑑 + �⃗⃗� • 𝑐 + �⃗⃗� • 𝑑
∎ �⃗� • �⃗⃗� = 0 (�⃗� ≠ 0, �⃗⃗� ≠ 0) ↔ �⃗� 𝑎𝑛𝑑 �⃗⃗� 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
For perpendicular vectors the dot/scalar product is 0.
The vector product of two vectors a × b Is a vector
The magnitude of the vector �⃗� × �⃗⃗� is equal to the area determined by both vectors.
|�⃗� × �⃗⃗�| = |�⃗�||�⃗⃗�| 𝑠𝑖𝑛 𝜃
Direction of the vector �⃗� × �⃗⃗� is given by right hand rule:
Point the fingers in direction of �⃗�; curl them toward �⃗⃗�. Your thumb points in the direction of cross product.
● DOT/SCALAR PRODUCT
● CROSS/VECTOR PRODUCT
4
Properties of vector product
∎ �⃗� × �⃗⃗� = − �⃗⃗� × �⃗�
∎ 𝑖𝑓 �⃗� 𝑎𝑛𝑑 �⃗⃗� 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟, 𝑡ℎ𝑒𝑛 |�⃗� × �⃗⃗�| = |�⃗�||�⃗⃗�|
∎ �⃗� × (�⃗⃗� + 𝑐) = �⃗� × �⃗⃗� + �⃗� × 𝑐
∎ (�⃗� + �⃗⃗�) × (𝑐 + 𝑑) = �⃗� × 𝑐 + �⃗� × 𝑑 + �⃗⃗� × 𝑐 + �⃗⃗� × 𝑑
∎ �⃗� × �⃗⃗� = 0 (�⃗� ≠ 0, �⃗⃗� ≠ 0) ↔ �⃗� 𝑎𝑛𝑑 �⃗⃗� 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
⇒ �⃗� × �⃗⃗� = (
𝑎𝑦𝑏𝑧 − 𝑎𝑧𝑏𝑦
𝑎𝑧𝑏𝑥 − 𝑎𝑥𝑏𝑧
𝑎𝑥𝑏𝑦 − 𝑎𝑦𝑏𝑥
) = |
𝑖̂ 𝑗̂ �̂�𝑎𝑥 𝑎𝑦 𝑎𝑧
𝑏𝑥 𝑏𝑦 𝑏𝑧
|
∎ �⃗� × �⃗⃗� = 0 (�⃗� ≠ 0, �⃗⃗� ≠ 0) ↔ �⃗� 𝑎𝑛𝑑 �⃗⃗� 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
∎ �⃗� × �⃗⃗� 𝑔𝑖𝑣𝑒𝑠 𝑎 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒𝑚
Find all vectors perpendicular to both �⃗� = (123
) 𝑎𝑛𝑑 �⃗⃗� = (321
)
�⃗� × �⃗⃗� = |𝑖̂ 𝑗̂ �̂�1 2 33 2 1
| = (−48
−4)
∴ 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑘(𝑖̂ − 2𝑗̂ + �̂�) 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑛𝑜𝑛 − 𝑧𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟.
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 vector 𝐴𝐵⃗⃗⃗⃗ ⃗⃗ = (3
−2−2
) and 𝐴𝐶⃗⃗⃗⃗⃗⃗ = (−105
)
1
2|𝐴𝐵⃗⃗⃗⃗ ⃗⃗ × 𝐵𝐶⃗⃗⃗⃗⃗⃗ | =
1
2||
𝑖̂ 𝑗̂ �̂�3 −2 −2
−1 0 5
|| ⤇1
2 |(
−10−13−2
)| =1
2 √(−10)2 + (−13)2 + (−2)2 = 8.26 𝑢𝑛𝑖𝑡𝑠2
�⃗� = 5𝑖̂ − 2𝑗̂ + �̂�
�⃗⃗� = 𝑖̂ + 𝑗̂ − 3�̂�
(a) Find the angle between them (b) Find the unit vector perpendicular to both
(a) 𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛 |�⃗⃗� × �⃗⃗�|
|�⃗⃗�||�⃗⃗�|
�⃗� × �⃗⃗� = |𝑖̂ 𝑗̂ �̂�5 −2 11 1 −3
| = 5𝑖̂ + 16𝑗̂ + 7�̂�
|�⃗� × �⃗⃗�| = |5𝑖̂ + 16𝑗̂ + 7�̂�| = √330
|�⃗� | = √30 |�⃗⃗� | = √11
𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛 1 = 𝜋/2
(b) �̂� = �⃗⃗� ×�⃗⃗�
|�⃗⃗� ×�⃗⃗�| =
1
√330 (
5167
)
5
Three points determine a plane. So, the fourth point is either on the plane (4 coplanar points) or not.
TEST: If the volume of the tetrahedron is zero points are coplanar.
● VOLUME of a PARALLELEPIPED
● VOLUME of a TETRAHEDRON
● TEST FOR FOUR COPLANAR POINTS
𝑉 = 𝑐 ● ( �⃗� × �⃗⃗�) = |
𝑐𝑥 𝑐𝑦 𝑐𝑧
𝑎𝑥 𝑎𝑦 𝑎𝑧
𝑏𝑥 𝑏𝑦 𝑏𝑧
| 𝑢𝑛𝑖𝑡𝑠3
𝑉 =1
6[𝑐 ● ( �⃗� × �⃗⃗�)] =
1
6 |
𝑐𝑥 𝑐𝑦 𝑐𝑧
𝑎𝑥 𝑎𝑦 𝑎𝑧
𝑏𝑥 𝑏𝑦 𝑏𝑧
| 𝑢𝑛𝑖𝑡𝑠3
Are the points A(1, 2, -4), B(3, 2, 0), C(2, 5, 1) and D(5, -3, -1) coplanar?
𝐴𝐵⃗⃗⃗⃗ ⃗⃗ = (204
) 𝐴𝐶⃗⃗⃗⃗⃗⃗ = (135
) 𝐴𝐷⃗⃗ ⃗⃗ ⃗⃗ = (4
−53
) ⤇
𝐴𝐵⃗⃗⃗⃗ ⃗⃗ ● [𝐴𝐶⃗⃗⃗⃗⃗⃗ × 𝐴𝐷⃗⃗ ⃗⃗ ⃗⃗ ] = |2 0 41 3 54 −5 3
| = 2(9 + 25) + 4(−5 − 12) = 0
∴ 𝐴, 𝐵, 𝐶 𝑎𝑛𝑑 𝐷 𝑎𝑟𝑒 𝑐𝑜𝑝𝑙𝑎𝑛𝑎𝑟 𝑄. 𝐸. 𝐷
6
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.
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.
The position vector �⃗⃗� of any general point P on the line passing
through point A and having direction vector �⃗⃗� is given by the equation
𝑟 = �⃗� + 𝐴𝑃⃗⃗⃗⃗⃗⃗ = �⃗� + 𝜆 �⃗⃗� 𝜆 ∈ 𝑅
where 𝜆 tells us how much of �⃗⃗� we need to take in order to get from A to P. (𝜆 = 3 for the particular P shown).
(𝑥𝑦𝑧
) = (
𝑎1
𝑎2
𝑎3
) + 𝜆 (
𝑏1
𝑏2
𝑏3
)
𝑟 = (𝑎1𝑖̂ + 𝑎2𝑗̂ + 𝑎3�̂�) + 𝜆(𝑏1�̂� + 𝑏2𝑗̂ + 𝑏3�̂�)
λ is called a parameter λ ∈ 𝑅
𝑥 = 𝑎1 + 𝜆𝑏1
𝑦 = 𝑎2 + 𝜆𝑏2
𝑧 = 𝑎3 + 𝜆𝑏3
𝑥−𝑎1
𝑏1 =
𝑦−𝑎2
𝑏2 =
𝑧−𝑎3
𝑏3 (= 𝜆)
● VECTOR EQUATION OF A LINE
● PARAMETRIC EQUATION OF A LINE
● CARTESIAN EQUATION OF A LINE
7
𝜃 = 𝑎𝑟𝑐 cos |�⃗⃗� • �⃗�
|�⃗⃗�||�⃗�||
Point P is at the shortest distance from the line when PQ is perpendicular to �⃗⃗�
𝑃𝑄⃗⃗ ⃗⃗ ⃗⃗ • �⃗⃗� = 0
● ANGLE BETWEEN TWO LINES
● SHORTEST DISTANCE FROM A POINT TO A LINE
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
4 5 9
5 2 3
AB
The Cartesian equation of this line is 3 5 2
1 9 3
x y z
(using the coordinates of point A).
The equivalent vector equation is 3 1
5 9
2 3
x
y t
z
.
acute angle !
𝜃 = 𝑎𝑟𝑐 cos�⃗⃗� • 𝑑
|�⃗⃗�||𝑑|
Find the shortest distance between 𝑟 = (131
) + 𝜆 (232
) and point P (1,2,3).
(The goal is to find Q first, and then |𝑃𝑄⃗⃗⃗⃗ ⃗⃗ |)
Point Q is on the line, hence its coordinates must satisfy line equation: (
𝑥𝑄
𝑦𝑄
𝑧𝑄
) = (1 + 2𝜆3 + 3𝜆1 + 2𝜆
) ⇒ 𝑃𝑄⃗⃗⃗⃗ ⃗⃗ = (2𝜆
1 + 3𝜆−2 + 2𝜆
)
(2𝜆
1 + 3𝜆−2 + 2𝜆
) • (232
) = 0 ⇒ 4𝜆 + 3 + 9𝜆 − 4 + 4𝜆 = 0 ⇒ 17 𝜆 = 1 ⇒ 𝜆 =1
17
𝑃𝑄⃗⃗⃗⃗ ⃗⃗ = (
2/1720/1732/17
) ⇒ 𝑓𝑖𝑛𝑑 |𝑃𝑄⃗⃗⃗⃗ ⃗⃗ |
8
To calculate the distance between two skew lines the lines are expressed using vectors,
: 𝑟 = �⃗� + 𝜆 �⃗⃗� 𝑎𝑛𝑑 𝑟 = 𝑐 + 𝜇 𝑑
The cross product of �⃗⃗⃗� and �⃗⃗⃗� is perpendicular to the lines, as is the unit vector:
�̂� =�⃗⃗� × 𝑑
|�⃗⃗� × 𝑑| 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒𝑠𝑒 𝑙𝑖𝑛𝑒𝑠
The distance between the lines is then
𝑑 = |�̂� • (𝑐 − �⃗�)| (sometimes I see it, sometimes I don’t)
● DISTANCE BTETWEEN TWO SKEW LINES
9
A plane is completely determined by two intersecting lines,
or
a fixed point A and two nonparallel direction vectors
The position vector �⃗⃗� of any general point P on the plane passing
through point A and having direction vectors �⃗⃗� and 𝑐 is given by the equation
𝑟 = �⃗� + 𝜆 �⃗⃗� + µ𝑐 𝜆, µ ∈ 𝑅 𝐴𝑃⃗⃗⃗⃗⃗⃗ = 𝜆 �⃗⃗� + µ𝑐
λ , μ are called a parameters λ,μ ∈ 𝑅
𝑥 = 𝑎1 + 𝜆𝑏1 + 𝜇𝑐1
𝑦 = 𝑎2 + 𝜆𝑏2 + 𝜇𝑐2
𝑧 = 𝑎3 + 𝜆𝑏3 + 𝜇𝑐3
The equation of a plane perpendicular to the vector �⃗⃗� = (
𝑛1
𝑛2
𝑛3
) and passing through the point 1 2 3( , , )a a a is
𝑟 • �⃗⃗� = �⃗� • �⃗⃗�
𝑛1𝑥 + 𝑛2𝑦 + 𝑛3𝑧 = 𝑑
𝑟 • �⃗⃗� = �⃗� • �⃗⃗� ⤇ 𝑛1𝑥 + 𝑛2𝑦 + 𝑛3𝑧 = 𝑛1𝑎1 + 𝑛2𝑎2 + 𝑛3𝑎3 = 𝑑
𝐷 = |𝑟 • �̂�| = |�⃗� • �̂�| = |�⃗� • �⃗⃗�
√𝑛12 + 𝑛2
2 + 𝑛32
| = |𝑛1𝑎1 + 𝑛2𝑎2 + 𝑛3𝑎3
√𝑛12 + 𝑛2
2 + 𝑛32
|
● VECTOR EQUATION OF A PLANE
● PARAMETRIC EQUATION OF A PLANE
● EQUATION OF A PLANE USING THE NORMAL VECTOR
● CARTESIAN EQUATION OF A PLANE
● DISTANCE FROM ORIGIN
�⃗⃗� • 𝐴𝑃⃗⃗⃗⃗⃗⃗ = 0 → �⃗⃗� • (𝑟 − �⃗�) = 0
10
GENERAL QUESTION:
What does the equation 3x + 4y = 12 give in 2 and 3 dimensions?
The angle between two planes is the same as the angle between the two lines carrying normal vectors
(meaning acute angle)
● THE ANGLE BETWEEN PLANE AND A LINE
● THE ANGLE BETWEEN TWO PLANES
𝑠𝑖𝑛 𝜃 = 𝑐𝑜𝑠 𝜙 = �⃗⃗� • 𝑑
|�⃗⃗�||𝑑|
𝜃 = 𝑎𝑟𝑐 𝑠𝑖𝑛 �⃗⃗⃗⃗� • �⃗⃗⃗⃗�
|�⃗⃗⃗⃗�| |�⃗⃗⃗⃗�|
𝑜𝑠 𝜃 = |�⃗⃗� • �⃗⃗⃗�|
|�⃗⃗�||�⃗⃗⃗�|
𝜃 = 𝑎𝑟𝑐 𝑐𝑜𝑠|�⃗⃗� • �⃗⃗⃗�|
|�⃗⃗�||�⃗⃗⃗�|
acute angle