Vectors Jeff Chastine

𝑉𝑒𝑐𝑡𝑜𝑟𝑠 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

𝑉𝑒𝑐𝑡𝑜𝑟𝑠 A vector is often denoted with an arrow above it (e.g. 𝑢 ) Row representation [x, y, z] (multiple columns) 15 −4 3 Column vector has multiple rows Vectors will be used in lighting equations 15 −4 3 Jeff Chastine

Example You Man-Bear-Pig How would I describe the 2D difference in location of a player and an enemy? You Man-Bear-Pig Jeff Chastine

Example How would I describe the 2D difference in location of a player and an enemy? (x2, y2) (x1, y1) Jeff Chastine

Example 𝑑𝑖𝑓𝑓 =[ 𝑥 1 − 𝑥 2 , 𝑦 1 − 𝑦 2 ] (x2, y2) (x1, y1) How would I describe the 2D difference in location of a player and an enemy? (x2, y2) (x1, y1) 𝑑𝑖𝑓𝑓 =[ 𝑥 1 − 𝑥 2 , 𝑦 1 − 𝑦 2 ] …or 𝑑𝑖𝑓𝑓=[∆𝑥, ∆𝑦] Note: a very useful 2D function is atan2 (∆y, ∆x) which gives you the angle!

Interpretation (x2, y2) (x1, y1) Magnitude (length) 𝑑𝑖𝑓𝑓 has both magnitude and direction (x2, y2) (x1, y1) Magnitude (length) Jeff Chastine

Interpretation (x2, y2) (x1, y1) (∆x, ∆y) Direction (0, 0) 𝑑𝑖𝑓𝑓 has both magnitude and direction (x2, y2) (x1, y1) (∆x, ∆y) (0, 0) Direction Jeff Chastine

Adding/Subtracting Vectors Do this component-wise Therefore, the vectors must be the same size Adding example 1 3 5 + 10 4 −3 = 11 7 2 Subtraction works the same way + = Jeff Chastine

Multiplication? Multiplying by a scalar (a single number) 6 * 1 3 5 = 6 18 30 What about multiplication? This isn’t really defined, but we do have Dot product Cross product Jeff Chastine

Magnitude and Normalization of Vectors Normalization is a fancy term for saying the vector should be of length 1 Magnitude is just its length and denoted 𝑉 Example for 1 3 5 mag = 𝑥 2 + 𝑦 2 + 𝑧 2 mag = 1 2 + 3 2 + 5 2 = 35 =~5.916079 To normalize the vector, divide each component by its magnitude Example from above 1 3 5 5.916079 = 0.169 0.50709 0.84515 Magnitude of 0.169 0.50709 0.84515 =~1 Jeff Chastine

The Dot Product 𝜃 Also called the inner or scalar product Multiply component-wise, then sum together Denoted using the dot operator 𝑢 ∙ 𝑣 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 Basis of almost all lighting calculations! 𝜃 Jeff Chastine

DOT PROduct Example 𝑣 𝜃 𝑢 Assume we have two vectors: 𝑢 = 1 0 0 𝑢 = 1 0 0 𝑣 = 0 1 0 These vectors are already normalized We expect the angle to be 90° Dot product is: 𝑢 ∙ 𝑣 = 1∗0 + 0∗1 + 0∗0 =0 acos 0 =90° 𝑣 𝜃 𝑢 Jeff Chastine

DOT PROduct Example 2 𝜃 𝑣 𝑢 Assume we have two vectors: 𝑢 = 1 0 0 𝑢 = 1 0 0 𝑣 = −1 0 0 We expect the angle to be 180° Dot product is: 𝑢 ∙ 𝑣 = 1∗−1 + 0∗0 + 0∗0 =−1 acos −1 =180° 𝜃 𝑣 𝑢 Jeff Chastine

DOT PROduct Example 3 𝑢 𝑣 Assume we have two vectors: 𝑢 = 1 0 0 𝑢 = 1 0 0 𝑣 = 1 0 0 We expect the angle to be 0° Dot product is: 𝑢 ∙ 𝑣 = 1∗1 + 0∗0 + 0∗0 =1 acos 1 =0° 𝑢 𝑣 Jeff Chastine

Projection 𝑉 𝛼 𝑊 Length of projection is: Can also be used to calculate the projection of one vector onto another 𝑉 𝛼 𝑊 Length of projection is: Then, multiply by normalized 𝑊/ 𝑊 𝑉 cos 𝛼 = 𝑉∙𝑊 𝑉 ∙ 𝑊 𝑝𝑟𝑜𝑗 𝑤 𝑉= 𝑉∙𝑊 𝑊 2 𝑊 Jeff Chastine

CROSS PRODUCT 𝑐 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Gives us a new vector that is perpendicular to the other two Denoted with the × operator Calculations: 𝑎 × 𝑏 =[ 𝑎 𝑦 𝑏 𝑧 − 𝑎 𝑧 𝑏 𝑦 𝑎 𝑧 𝑏 𝑥 − 𝑎 𝑥 𝑏 𝑧 𝑎 𝑥 𝑏 𝑦 − 𝑎 𝑦 𝑏 𝑥 ] Interesting: If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 =− 𝑐 The magnitude (length) of the new vector is the sine of the angle (if normalized) 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Jeff Chastine

CROSS PRODUCT 𝑐 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Gives us a new vector that is perpendicular to the other two Denoted with the × operator Calculations: 𝑎 × 𝑏 =[ 𝑎 𝑦 𝑏 𝑧 − 𝑎 𝑧 𝑏 𝑦 𝑎 𝑧 𝑏 𝑥 − 𝑎 𝑥 𝑏 𝑧 𝑎 𝑥 𝑏 𝑦 − 𝑎 𝑦 𝑏 𝑥 ] Interesting: If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 =− 𝑐 The magnitude (length) of the new vector is the sine of the angle (if normalized) 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Jeff Chastine

CROSS PRODUCT 𝑐 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Gives us a new vector that is perpendicular to the other two Denoted with the × operator Calculations: 𝑎 × 𝑏 =[ 𝑎 𝑦 𝑏 𝑧 − 𝑎 𝑧 𝑏 𝑦 𝑎 𝑧 𝑏 𝑥 − 𝑎 𝑥 𝑏 𝑧 𝑎 𝑥 𝑏 𝑦 − 𝑎 𝑦 𝑏 𝑥 ] Interesting: If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 =− 𝑐 The magnitude (length) of the new vector is the sine of the angle (if normalized) 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Jeff Chastine

CROSS PRODUCT 𝑐 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Gives us a new vector that is perpendicular to the other two Denoted with the × operator Calculations: 𝑎 × 𝑏 =[ 𝑎 𝑦 𝑏 𝑧 − 𝑎 𝑧 𝑏 𝑦 𝑎 𝑧 𝑏 𝑥 − 𝑎 𝑥 𝑏 𝑧 𝑎 𝑥 𝑏 𝑦 − 𝑎 𝑦 𝑏 𝑥 ] Interesting: If 𝑎 × 𝑏 = 𝑐 then 𝑏 × 𝑎 =− 𝑐 The magnitude (length) of the new vector is the sine of the angle (if normalized) 𝑎 𝜃 𝑏 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑏𝑥 𝑏𝑦 𝑏𝑧 Jeff Chastine

Fun Questions How do we find the normal of a triangle? How can we determine if a polygon is facing away from the camera? Jeff Chastine

Fun Questions P0 P1 P2 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 Jeff Chastine

Fun Questions P0 𝑢 𝑣 P1 P2 Make some vectors… How do we find the normal of a triangle? How can we determine if a polygon is facing away from the camera? P0 𝑢 𝑣 P1 P2 Make some vectors… Jeff Chastine

Fun Questions P0 𝑢 𝑣 P1 P2 Make some vectors… How do we find the normal of a triangle? How can we determine if a polygon is facing away from the camera? P0 𝑢 𝑣 P1 P2 Make some vectors… Jeff Chastine

Fun Questions 𝑢 × 𝑣 P0 𝑢 𝑣 P1 P2 Take the cross product 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 𝑣 P1 P2 Take the cross product Jeff Chastine

Fun Questions 𝑁 P0 P1 P2 How do we find the normal of a triangle? How can we determine if a polygon is facing away from the camera? 𝑁 P0 P1 P2 Jeff Chastine

Fun Questions 𝑁 P0 P1 P2 How do we find the normal of a triangle? How can we determine if a polygon is facing away from the camera? 𝑁 P0 P1 P2 Jeff Chastine

Fun Questions 𝑁 P0 camera P1 P2 How do we find the normal of a triangle? How can we determine if a polygon is facing away from the camera? 𝑁 P0 camera P1 P2 Jeff Chastine

Fun Questions 𝑁 P0 𝑐𝑎𝑚 camera P1 P2 How do we find the normal of a triangle? How can we determine if a polygon is facing away from the camera? 𝑁 P0 𝑐𝑎𝑚 camera P1 P2 Jeff Chastine

Fun Questions 𝐼𝑓 acos 𝑁∙𝑐𝑎𝑚 <90°, 𝑖 𝑡 ′ 𝑠 𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑁 P0 𝑐𝑎𝑚 camera 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°, 𝑖 𝑡 ′ 𝑠 𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑁 P0 𝑐𝑎𝑚 camera P1 P2 Jeff Chastine Assuming 𝑁 and 𝑐𝑎𝑚 are normalized

A final Note Can multiply a matrix and vector to: Rotate the vector Translate the vector Scale the vector Etc.. This operation returns a new vector Jeff Chastine

The End Image of a triangle facing away from the camera Jeff Chastine