VECTORS
JEFF CHASTINE
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𝑉𝑒𝑐𝑡𝑜𝑟𝑠
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A mathematical structure that has more than one “part” (e.g. an array)
• 2D vectors might have x and y
• 3D vectors might have x, y and z
• 4D vectors might have x, y, z and w
•
Vectors can represent a point in space
•
Vectors commonly represent both:
• Direction
• Magnitude
JEFF CHASTINE
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𝑉𝑒𝑐𝑡𝑜𝑟𝑠
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A vector is often denoted with an arrow above it (e.g.𝑢)
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Row representation [x, y, z] (multiple columns)
15 − 4 3
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Column vector has multiple rows
15
−4
3
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Vectors will be used in lighting equations
JEFF CHASTINE
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EXAMPLE
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How would I describe the 2D difference in location of a player and an enemy?
You
Man-Bear-Pig
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EXAMPLE
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How would I describe the 2D difference in location of a player and an enemy?
(x2, y2)
(x1, y1)
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EXAMPLE
•
How would I describe the 2D difference in location of a player and an enemy?
(x2, y2)
(x1, y1)
𝑑𝑖𝑓𝑓 = [ 𝑥1 − 𝑥2 , 𝑦1 − 𝑦2 ]
…or 𝑑𝑖𝑓𝑓 = [∆𝑥, ∆𝑦]
Note: a very useful 2D function is atan2 (∆y, ∆x) which gives you the angle!
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INTERPRETATION
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𝑑𝑖𝑓𝑓 has both magnitude and direction
(x2, y2)
(x1, y1)
Magnitude (length)
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INTERPRETATION
•
𝑑𝑖𝑓𝑓 has both magnitude and direction
(x2, y2)
(x1, y1)
(∆x, ∆y)
(0, 0)
JEFF CHASTINE
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ADDING/SUBTRACTING VECTORS
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Do this component-wise
•
Therefore, the vectors must be the same size
•
Adding example
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1 3 5 + 10 4 − 3 = 11 7 2
• Subtraction works the same way
+
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=
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MULTIPLICATION?
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Multiplying by a scalar (a single number)
• 6 * 1 3 5 = 6 18 30
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What about multiplication?
• This isn’t really defined, but we do have
• Dot product
• Cross product
JEFF CHASTINE
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MAGNITUDE AND NORMALIZATION OF VECTORS
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Normalization is a fancy term for saying the vector should be of length 1
•
Magnitude is just its length and denoted 𝑉
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Example for 1 3 5
• mag = 𝑥 2 + 𝑦 2 + 𝑧 2
• mag = 12 + 32 + 52 = 35 = ~5.916079
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To normalize the vector, divide each component by its magnitude
•
Example from above
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1 3 5
5.916079
= 0.169 0.50709 0.84515
• Magnitude of 0.169 0.50709 0.84515 = ~1
JEFF CHASTINE
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THE DOT PRODUCT
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Also called the inner or scalar product
•
Multiply component-wise, then sum together
•
Denoted using the dot operator 𝑢 ∙ 𝑣
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Example
•
•
1 2 4 ∙ 8 1 − 2 = 1 ∗ 8 + 2 ∗ 1 + 4 ∗ −2 = 2.0
Why is this so cool?
• If normalized, it’s the cosine of the angle θ between
the two vectors!
• Use acos(𝜃) to “undo” that
𝜃
• Basis of almost all lighting calculations!
JEFF CHASTINE
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DOT PRODUCT EXAMPLE
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Assume we have two vectors:
𝑣
• 𝑢= 1 0 0
• 𝑣= 0 1 0
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These vectors are already normalized
•
We expect the angle to be 90°
•
Dot product is:
𝜃
𝑢
• 𝑢∙𝑣 = 1∗0 + 0∗1 + 0∗0 =0
• acos 0 = 90°
JEFF CHASTINE
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DOT PRODUCT EXAMPLE 2
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Assume we have two vectors:
• 𝑢= 1 0 0
• 𝑣 = −1 0 0
•
•
We expect the angle to be 180°
Dot product is:
𝜃
𝑣
𝑢
• 𝑢 ∙ 𝑣 = 1 ∗ −1 + 0 ∗ 0 + 0 ∗ 0 = −1
• acos −1 = 180°
JEFF CHASTINE
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DOT PRODUCT EXAMPLE 3
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Assume we have two vectors:
• 𝑢= 1 0 0
• 𝑣= 1 0 0
•
We expect the angle to be 0°
•
Dot product is:
𝑢 𝑣
• 𝑢∙𝑣 = 1∗1 + 0∗0 + 0∗0 =1
• acos 1 = 0°
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PROJECTION
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Can also be used to calculate the projection of one vector onto another
𝑉
𝛼
Length of projection is:
𝑉∙𝑊
𝑉 cos 𝛼 =
𝑊
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𝑊
Then, multiply by normalized 𝑊/ 𝑊
𝑝𝑟𝑜𝑗𝑤 𝑉 =
𝑉∙𝑊
𝑊
𝑊 2
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𝑐
CROSS PRODUCT
•
Gives us a new vector that is perpendicular to the other two
•
Denoted with the × operator
•
Calculations:
𝑎
𝜃
𝑎 × 𝑏 = [𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦
𝑎𝑥
𝑏𝑥
•
𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧
𝑎𝑦
𝑏𝑦
𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 ]
𝑏
𝑎𝑧
𝑏𝑧
Interesting:
• If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 = −𝑐
• The magnitude (length) of the new vector is the sine of the angle (if normalized)
JEFF CHASTINE
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𝑐
CROSS PRODUCT
•
Gives us a new vector that is perpendicular to the other two
•
Denoted with the × operator
•
Calculations:
𝑎
𝜃
𝑎 × 𝑏 = [𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦
𝑎𝑥
𝑏𝑥
•
𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧
𝑎𝑦
𝑏𝑦
𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 ]
𝑏
𝑎𝑧
𝑏𝑧
Interesting:
• If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 = −𝑐
• The magnitude (length) of the new vector is the sine of the angle (if normalized)
JEFF CHASTINE
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𝑐
CROSS PRODUCT
•
Gives us a new vector that is perpendicular to the other two
•
Denoted with the × operator
•
Calculations:
𝑎
𝜃
𝑎 × 𝑏 = [𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦
𝑎𝑥
𝑏𝑥
•
𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧
𝑎𝑦
𝑏𝑦
𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 ]
𝑏
𝑎𝑧
𝑏𝑧
Interesting:
• If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 = −𝑐
• The magnitude (length) of the new vector is the sine of the angle (if normalized)
JEFF CHASTINE
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𝑐
CROSS PRODUCT
•
Gives us a new vector that is perpendicular to the other two
•
Denoted with the × operator
•
Calculations:
𝑎
𝜃
𝑎 × 𝑏 = [𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦
𝑎𝑥
𝑏𝑥
•
𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧
𝑎𝑦
𝑏𝑦
𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 ]
𝑏
𝑎𝑧
𝑏𝑧
Interesting:
• If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 = −𝑐
• The magnitude (length) of the new vector is the sine of the angle (if normalized)
JEFF CHASTINE
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FUN QUESTIONS
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How do we find the normal of a triangle?
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How can we determine if a polygon is facing away from the camera?
JEFF CHASTINE
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FUN QUESTIONS
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How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
P0
P1
P2
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
𝑢
P1
P0
𝑣
P2
Make some vectors…
JEFF CHASTINE
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
P0
𝑢
P1
𝑣
P2
Make some vectors…
JEFF CHASTINE
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
𝑢 × 𝑣
P0
𝑢
P1
𝑣
P2
Take the cross product
JEFF CHASTINE
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
P0
𝑁
P1
P2
JEFF CHASTINE
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
P0
𝑁
P1
P2
JEFF CHASTINE
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
P0
𝑁
camera
P1
P2
JEFF CHASTINE
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
P0
𝑁
camera
P1
P2
JEFF CHASTINE
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FUN QUESTIONS
•
How do we find the normal of a triangle?
•
How can we determine if a polygon is facing away from the camera?
𝐼𝑓 acos 𝑁 ∙ 𝑐𝑎𝑚 < 90°, 𝑖𝑡 ′ 𝑠 𝑣𝑖𝑠𝑖𝑏𝑙𝑒
P0
𝑁
camera
P1
P2
JEFF CHASTINE
Assuming 𝑁 and 𝑐𝑎𝑚 are normalized
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A FINAL NOTE
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Can multiply a matrix and vector to:
• Rotate the vector
• Translate the vector
• Scale the vector
• Etc..
•
This operation returns a new vector
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THE END
Image of a triangle facing away from the camera
JEFF CHASTINE
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