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