Dot Product

Definition of dot product in polar coordinates is presented in section "Dot product in coordinate systems" and Example 23.    ■

1. Any nonzero vector will work; for instance, v = (3, −2, 1) ∈ ℝ³. Then \[ \mathbf{v} \bullet \mathbf{v} = 3^2 + (-2)^2 + 1^2 = 9+4+1 = 14 > 0. \]
         {3, -2, 1} . {3, -2, 1}
         14
2. Commutativity holds because the dot product is implemented element- wise, and each element-wise multiplication is simply the product of two scalars. Scalar multiplication is commutative, and therefore the dot product is commutative.

   Let \[ \mathbf{v} = \left( 1, 2, 3 \right) , \quad \mathbf{u} = \left( 4, -6, 5 \right) \in \mathbb{R}^3 . \] Then their scalar product is 7, independently of the order of multiplication,as Mathematica confirms:

         v = {1, 2, 3}; u = {4, -6, 5}; v.u
         7
         Dot[u, v]
         7
3. Suppose we need to find a dot product of two numerical vectors v • u, one of which has large entries. For instance, \[ \mathbf{v} = \begin{pmatrix} 3791 \\ -5688 \\ 2894 \end{pmatrix} , \quad \mathbf{u} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} . \] Scalar product of these numerical vectors involves large unpleasant multiplications and summation. Using distributive property, we break vector v into sum of four vectors: \[ \mathbf{v} = \mathbf{v}_1 + \mathbf{v}_2 + \mathbf{v}_3 + \mathbf{v}_4 , \] where \[ \mathbf{v}_1 = \begin{pmatrix} 1 \\ -8 \\ 4 \end{pmatrix} , \ \mathbf{v}_2 = \begin{pmatrix} 90 \\ -80 \\ 90 \end{pmatrix} , \ \mathbf{v}_3 = \begin{pmatrix} 700 \\ -600 \\ 800 \end{pmatrix} , \ \mathbf{v}_4 = \begin{pmatrix} 3000 \\ -5000 \\ 2000 \end{pmatrix} . \] The corresponding four scalar products are not tedious to find:
         u = {3,2,4}; v1 = {1,-8,4}; v2 = {90, -80, 90}; v3 = {700,-600,800}; v4 = {3000,-5000,2000}; d1 = u.v1
         3
         d2 = u.v2
         470
         d3 = u.v3
          4100
         d4 = u.v4
          7000
   Adding these four numbers, we get the required dot product: \begin{align*} \mathbf{v} \bullet \mathbf{u} &= \left( \mathbf{v}_1 + \mathbf{v}_2 + \mathbf{v}_3 + \mathbf{v}_4 \right) \bullet \mathbf{u} = \mathbf{v}_1 \bullet \mathbf{u} + \mathbf{v}_2 \bullet \mathbf{u} + \mathbf{v}_3 \bullet \mathbf{u} + \mathbf{v}_4 \bullet \mathbf{u} \\ &= 3+470+4100+7000 = 11573 . \end{align*}
         d1+d2+d3+d4
         11573
4. We set λ = 3.1415926, v = (236, -718), u = (892, 435). Without computer assistance determination of corresponding dot products will be time consuming. So we ask Mathematica for help and find that \[ \left( \lambda\mathbf{u} \right) \bullet \mathbf{v} = \lambda \left( \mathbf{u} \bullet \mathbf{v} \right) = \mathbf{u} \bullet \left( \lambda \mathbf{v} \right) \approx 319871. \]
         la =3.1415926; v = {236,-718}; u = {892, 435}; (la*u) . v
         319871.
         u . (la*v)
         319871.
         la*(v . u)
         319871.
5. Let us take a singular matrix and two 3-column vectors: \[ \mathbf{A} = \begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix} , \quad \mathbf{u} = \begin{pmatrix} 35 \\ -11 \\ 17 \end{pmatrix} , \quad \mathbf{v} = \begin{pmatrix} 23 \\ 97 \\ 41 \end{pmatrix} . \]
         A = {{1, 2, 3}, {4, 5, 6}, {7,8 ,9}};
   u = {35,-11,17}; v = {23,97,41};
   We set vectors v and u as 3-tuples (elements of ℝ³) but not column vectors (elements of ℝ^3×1) because Mathematica is smart to understand that we want column vectors.
          u = {{35}, {-11}, {17}}; v = {{23}, {97}, {41}};
         Dot[u, A . v]
          25049
   and
         Dot[Transpose[A] . u, v]
          25049
   Now we check the same property for row vectors:
          Dot[u . A, v]
          25049
          Dot[v . Transpose[A], u]
          25049

   Rectangular matrix:    we consider 3-by-2 matrix \[ \mathbf{A} = \begin{bmatrix} 1&2 \\ 3&4 \\ 5&6 \end{bmatrix} \] and two column vectors \[ \mathbf{u} = \begin{pmatrix} a \\ b \end{pmatrix}, \qquad \mathbf{v} = \begin{pmatrix} c \\ d \\ e \end{pmatrix} . \] Then \[ \mathbf{A}\,\mathbf{u} = \begin{bmatrix} 1&2 \\ 3&4 \\ 5&6 \end{bmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 1\cdot a + 2 \cdot b \\ 3 \cdot a + 4 \cdot b \\ 5 \cdot a + 6 \cdot b \end{pmatrix} . \] Its dot product with v becomes \[ \mathbf{v} \bullet \mathbf{A}\,\mathbf{u} = c \left( a + 2\,b \right) + d \left( 3\,a + 4\, b \right) + e \left( 5\,a + 6\, b \right) . \tag{A} \] On the other hand, \[ \mathbf{A}^{\mathrm T} \mathbf{v} = \begin{bmatrix} 1 & 3 & 5 \\ 2&4&6 \end{bmatrix} \begin{pmatrix} c \\ d \\ e \end{pmatrix} = \begin{pmatrix} 1\cdot c + 3\cdot d + 5\cdot e \\ 2 \cdot c + 4 \cdot d + 6 \cdot e \end{pmatrix} . \] Then its dot product with u becomes \[ \mathbf{A}^{\mathrm T} \mathbf{v} \bullet \mathbf{u} = a \left( c + 3\, d + 5\, e \right) + b \left( 2\,c + 4\, d + 6\, e \right) . \tag{B} \] Do you need a computer assistance to check that expression (A) is the same as (B) ? I don't need it.

6. Extra warning:    We demonstrate that the identity   v • (u • w) = (v • u) • w is not valid for arbitrary vectors v, u, and w. It is true only when v is a scalar multiple of w. Indeed, let k = u • w and c = v • u. Then \[ \mathbf{v} \bullet \left( \mathbf{u} \bullet \mathbf{w} \right) = \left( \mathbf{v} \bullet \mathbf{u} \right) \bullet \mathbf{w} \quad \iff \quad \mathbf{v} k = c \mathbf{w} . \] from the latter, it follows that vectors v and w are collinear.

   We choose three vectors (and write them in column form): \[ \mathbf{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} , \quad \mathbf{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} , \quad \mathbf{w} = \begin{pmatrix} 5 \\ 6 \end{pmatrix} . \] Then \[ \mathbf{v} \bullet \left( \mathbf{u} \bullet \mathbf{w} \right) = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \cdot 39 = \begin{pmatrix} 39 \\ 78 \end{pmatrix} \]

         v = {1, 2}; u = {3, 4}; w = {5, 6};
         u . w
         39
         v*39
         {39, 78}
   and \[ \left( \mathbf{v} \bullet \mathbf{u} \right) \bullet \mathbf{w} = 11 \cdot \begin{pmatrix} 5 \\ 6 \end{pmatrix} = \begin{pmatrix} 55 \\ 66 \end{pmatrix} . \]
         Dot[v, u]
          11
         11*w
         {55, 66}
   However, for some vectors, we observe v • (u • w)) = (v • u) • w. Let us pick up three vectors \[ \mathbf{v} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} , \quad \mathbf{u} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} , \quad \mathbf{w} = \begin{pmatrix} 2 \\ 4 \end{pmatrix} . \] Then \[ \mathbf{v} \bullet \left( \mathbf{u} \bullet \mathbf{w} \right) = \mathbf{v} \,26 = 26 \begin{pmatrix} 1 \\ 2 \end{pmatrix} , \]
          v = {1, 2}; u = {3,5}; w = {2,4}; u.w
          26
   and \[ \left( \mathbf{v} \bullet \mathbf{u} \right) \bullet \mathbf{w} = 13 \mathbf{w} = 13 \begin{pmatrix} 2 \\ 4 \end{pmatrix} . \]
          v.u
          13
   Since \[ 26 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = 13 \begin{pmatrix} 2 \\ 4 \end{pmatrix} , \] we conclude that this identity is valid for collinear vectors v, and w, and arbitrary u.

Now we set vector u to be u = 2.71828 v. Using Mathematica, we repeat all previous calculations for these two vectors.

Equality:    We consider two vectors in ℝ²: \[ \mathbf{v} = \left( 1 , 2 \right) , \qquad \mathbf{u} = \left( 2 , 4 \right) \] They are linearly dependent because u = 2v. Their scalar product is \[ \left( 1 , 2 \right) \bullet \left( 2 , 4 \right) = 1 \cdot 2 + 2 \cdot 4 = 10. \] I hope that you can square this number without a computer assistance. Norms squared of these vectors are \[ \| \mathbf{v} \|^2 = 1^2 + 2^2 = 5 , \qquad \| \mathbf{u} \|^2 = 2^2 + 4^2 = 4 + 16 = 20 . \] Their product becomes \[ \| \mathbf{v} \|^2 \cdot \| \mathbf{u} \|^2 = 5 \cdot 20 = 100 = 10^2 = \left( \mathbf{v} \bullet \mathbf{u} \right)^2 . \]    ■

Another matrix:    We repeat all previous calculations with another not positive definite matrix \[ \mathbf{A} = \begin{bmatrix} -249& -292& -276& 412 \\ 136& 161& 150& -226 \\ 185& 218& 205& -308 \\ 69& 82& 76& -115 \end{bmatrix} . \] Its characteristic polynomial is det(λI − A) = (λ − 1)³(λ + 1). The corresponding dot products are

A first relation between the angles α and β is found if we project the sides AC and BC onto AB \[ c = b\,\cos\alpha + a\,\cos\beta . \] Projecting similarly onto the other sides, we are led to two further equations (which can be obtained from the first one by the use of circular permutations). Here is the system \[ \begin{cases} b\,\cos\alpha + a\,\cos\beta \phantom{+b\,\cos\gamma} &= c , \\ \phantom{b\,\cos\alpha} c \,\cos\beta + b\,\cos\gamma &= a , \\ c\,\cos\alpha + \phantom{c\.\cos\beta} + a\,\cos\gamma &= b . \end{cases} \] It is linear in the three variables x₁ = cosα, x₂ = cosβ, x₃ = cosγ. Let us solve this system by Gaussian elimination: \begin{align*} \left( \begin{array}{ccc|l} b&a&0&c \\ 0&c&b&a \\ c&0&a&b \end{array} \right) &\sim \left( \begin{array}{ccc|l} b&a&0&c \\ 0&c&b&a \\ 0& -\frac{c}{b}\,a & a& b - \frac{c}{b}\,c \end{array} \right) \\ &\sim \left( \begin{array}{ccc|l} b&a&0&c \\ 0&c&b&a \\ 0&0& 2a & b - \frac{c^2}{b} + \frac{a^2}{b} \end{array} \right) \end{align*} This yields

Using Mathematica, we verify properties included in Theorem 6 with the following examples.

1. This formulas follows from the identity: \[ \mathbf{v} \bullet \mathbf{v} = v_1^2 + v_2^2 + \cdots + v_n^2 . \] which is zero only when all components of vector v are zeroes.
2. We generate a vector of size five:
        v = Flatten[RandomReal[{-1, 1}, {1, 5}]]
        {0.371645, -0.811594, 0.591386, -0.037053, 0.758338}
   Upon choosing a real scalar λ = 2.71, we multiply
        2.71*v
        {1.00716, -2.19942, 1.60266, -0.100414, 2.0551}
   Its norm is
        Norm[2.71*v]
        3.55722
   Now we calculate the norm of v and multiply the result by λ = 2.71,
        2.71*Norm[v]
        3.55722
3. We generate two vectors of length five:
        v = RandomInteger[{-8, 8}, {1, 5}]
        {{-1, -7, 0, 6, 1}}
        u = RandomInteger[{-7, 8}, {1, 5}]
        {{-1, 3, -2, 2, 3}}
   In order to verify the triangle inequality, we evaluate both sides:
        N[Norm[u] + Norm[v]]
        14.5235
        N[Norm[u + v]]
        10.198
   \[ 10.198 \approx \| \mathbf{v} + \mathbf{u} \| < \| \mathbf{v} \| + \| \mathbf{u} \| \approx 14.5235 . \]
4. We choose randomly two vectors
        v = Flatten[RandomInteger[{-7, 9}, {1, 6}]]
       u = Flatten[RandomInteger[{-7, 9}, {1, 6}]]
        {5, 9, -5, 1, 2, -2}
       {-1, -6, 3, -5, -3, -1}
   Their norms are
        Norm[v]
        2 Sqrt[35]
        Norm[u]
        9
   The difference of norms is
        N[Norm[v] - Norm[u]]
       2.83216
   The norm of their difference is
        N[Norm[v - u]]
        19.6723
   The sum of norms is
        N[Norm[v] + Norm[u]]
        20.8322
   These numbers confirm inequalities in part 4.

Using points A(0, 0, c), B(𝑎, 0, 0), and C(0, b, 0), we find vectors \[ \mathbf{u} = \vec{BA} = (-a, 0, c) \qquad\mbox{and} \qquad \mathbf{v} = \vec{BC} = (-a, b, 0) . \] Their norms \[ \| \mathbf{u} \| = \sqrt{a^2 + c^2} , \qquad \|\mathbf{v} \| = \sqrt{a^2 + b^2} \] and dot product \[ \mathbf{u} \bullet \mathbf{v} = a^2 \] show that formula (8) can be written as \[ \frac{\mathbf{u} \bullet \mathbf{v}}{\| \mathbf{u} \| \cdot \| \mathbf{v} \|} = \frac{a^2}{\sqrt{a^2 + c^2}\cdot\sqrt{a^2 + b^2}} = \cos\theta \] because the ratio is definitely less than 1. For instance, if 𝑎 =1, b = 2, and c = 3, we get \[ \cos\theta = \frac{1}{\sqrt{5}\cdot \sqrt{10}} \qquad \Longrightarrow \qquad \theta = \mbox{arctan}\left( \frac{1}{5\sqrt{2}} \right) \approx 0.14049 . \]

Another angle problem:    We consider the same cuboid, but now we are going to determine the angle between a diagonal on a face and a diagonal of the cuboid. Upon plotting Figure 16.2, we are after the angle ∠ABD.

Since point D has coordinates (0, b, c), we find vector \( \displaystyle \quad \mathbf{v} = \vec{BD} = \left( -a, b, c \right) . \quad \) Then the dot product of vectors u and v becomes \[ \mathbf{u} \bullet \mathbf{v} = \left( -a, 0, c \right) \bullet \left( -a, b, c \right) = a^2 + c^2 . \] Then formula (8) yields \[ \cos\theta = \frac{\mathbf{u} \bullet \mathbf{v}}{\| \mathbf{u} \| \, \| \mathbf{v} \|} = \frac{a^2 + c^2}{\sqrt{a^2 + c^2}\,\sqrt{a^2 + b^2 + c^2}} . \] Taking inverse cosine function, we find the angle to be \[ \theta = \mbox{arccos} \left( \frac{\sqrt{a^2 + c^2}}{\sqrt{a^2 + b^2 + c^2}} \right) . \] For 𝑎 =1, b = 2, c = 3, we get \[ \theta = \mbox{arccos} \left( \frac{\sqrt{10}}{\sqrt{14}} \right) \approx 0.563943 = 32.3115^{\circ} . \]

Another basis:    \[ \beta = \left[ \mathbf{b}_1 , \mathbf{b}_2 \right] = \left[ \mathbf{i} , \mathbf{i} + \mathbf{j} \right] . \] The vector v = (−3, 4) has the following coordinates in basis β: \[ \mathbf{v} = \left( -3, 4 \right) \qquad \Longrightarrow \qquad \left[\!\left[ \mathbf{v} \right]\!\right]_{\beta} = \left( v^1 , v^2 \right) = \left( -7 , 4 \right) \] because \[ \mathbf{v} = -3\mathbf{i} + 4\mathbf{j} = -3\mathbf{i} + 4 \left( \mathbf{i} + \mathbf{j} - \mathbf{i} \right) = -3\mathbf{i} + 4 \left( \mathbf{i} + \mathbf{j} \right) - 4\mathbf{i} . \] Here we used a mathematical trick: add and subtruct the same value. The second component j is replaced by j + i − i = b₂ − i. The dot product in basis β becomes \begin{align*} \mathbf{v} \bullet \mathbf{v} &= g_{1,1} v^1 v^1 + g_{1,2} v^1 v^2 + g_{2,1} v^2 v^1 + g_{2,2} v^2 v^2 \\ &= g_{1,1} \left( -7 \right)^2 + g_{1,2} \left( -7 \right) \left( 4 \right) + g_{2,1} \left( 4 \right) \left( -7 \right) + g_{2,2} \left( 4 \right)^2 \\ &= g_{1,1} 49 - g_{1,2} 28 - g_{2,1} 28 + g_{2,2} 16 . \end{align*} The components of the metric tensor are \begin{align*} g_{1,1} &= \mathbf{b}^1 \bullet \mathbf{b}^1 = \mathbf{i} \bullet \mathbf{i} = 1 , \\ g_{1,2} &= \mathbf{b}^1 \bullet \mathbf{b}^2 = \mathbf{i} \bullet \left( \mathbf{i} + \mathbf{j} \right) = \mathbf{i} \bullet \mathbf{i} + \mathbf{i} \bullet \mathbf{j} = 1 + 0 = 1 , \\ g_{2,1} &= \mathbf{b}^2 \bullet \mathbf{b}^1 = \mathbf{i} \bullet \left( \mathbf{i} + \mathbf{j} \right) = \mathbf{i} \bullet \mathbf{i} + \mathbf{i} \bullet \mathbf{j} = 1 + 0 = 1 , \\ g_{2,2} &= \mathbf{b}^2 \bullet \mathbf{b}^2 = \left( \mathbf{i} + \mathbf{j} \right) \bullet \left( \mathbf{i} + \mathbf{j} \right) = \mathbf{i} \bullet \mathbf{i} + \mathbf{j} \bullet \mathbf{j} = 2 . \end{align*} Using these values, we calculate the dot product \begin{align*} \mathbf{v} \bullet \mathbf{v} &= 49 - 28 - 26 + 2\cdot 16 = 25 . \end{align*}

Naturally, other physical quantities can be expressed in such a way. For example, the electrostatic potential energy gained by moving a charge q along a path C in an electric field E is     −q∫_C E • dr. We may also note that Ampere's law concerning the magnetic field B associated with a current-carrying wire can be written as \[ \oint_C {\bf B} \bullet {\text d}{\bf r} = \mu_0 I , \] where I is the current enclosed by a closed path C traversed in a right-handed sense with respect to the current direction.    ■

2D case:    we start with flat (ℝ²).

The area of a 2D triangle whose vertices are 𝑎 = (x_𝑎, y_𝑎), b = (x_b, y_b), c = (x_c, y_c) (as shown in figure 1) is given by \[ \mbox{area} = \frac{1}{2} \begin{vmatrix} x_b - x_a & x_c - x_a \\ y_b - y_a & y_c - y_a \end{vmatrix} . \]

Barycentric coordinates allow us to express the coordinates of p = (x, y) in terms of 𝑎, b, c. More specifically, the barycentric coordinates of p are the numbers β and γ such that \[ p = a + \beta\left( b-a \right) + \gamma \left( c -a \right) \] If we regroup 𝑎, b, c, we obtain \begin{align*} p &= a + \beta\,b - \beta\,a + \gamma\,c - \gamma\,a \\ &= \left( a - \beta \right) a + \beta\,b + \gamma\,c . \end{align*} It is customary to define a third variable α by \[ \alpha = 1 - \beta - \gamma . \] Then we have \[ p = \alpha\,a + \beta\, b + \gamma\,c , \qquad \alpha + \beta + \gamma = 1 . \] The barycentric coordinates of the point p in terms of the points 𝑎, b, c are the numbers α, β, γ such that p = α𝑎 + βb + γc, with α + β + γ = 1.

Barycentric coordinate are defined for all points in the plane. They have several nice features:

1. A point p is inside the triangle defined by 𝑎, b, c if and only if \[ 0 < \alpha < 1 , \quad 0 < \beta < 1 , \quad 0 < \gamma < 1 . \] This property provides an easy way to test if a point is inside a triangle.
2. If one of the barycentric coordinates is 0 and the other two are between 0 and 1, the corresponding point p is on one of the edges of the triangle.
3. If any barycentric coordinate is less than zero, then p must lie outside of the triangle.
4. If two of the barycentric coordinates are zero and the third is 1, the point p is at one of the vertices of the triangle.
5. By changing the values of α, β, γ between 0 and 1, the point p will move smoothly inside the triangle. This can (and will) be applied to other properties of the vertices such as color problem.
6. The center of the triangle is obtained when α = β = γ = ⅓. If the triangle is made of a certain substance which is evenly distributed throughout the triangle, then these values of α, β, γ would give us the center of gravity.

Let A_𝑎, A_b and A_c be as in figure 1 and let A denote the area of the triangle. Also note that the point inside the triangle on figure 1 is the point we called p. Consider the triangles in the figure. These are different triangles drawn for a fixed value of β. They have the same area since they have the same base and height. This area was denoted A_b on figure 1. Thus, we see that A_b only depends on β. Therefore, we have \[ A_b = C\beta . \] for some constant C. When p is on b that is when β = 1, we have A_b = A. Hence, A = C. Therefore we see that \[ \beta = \frac{A_b}{A} . \] Similarly, we have \[ \alpha = \frac{A_a}{A} , \qquad \gamma = \frac{A_c}{A} . \] In coordinates, these parameters become \[ \beta = \frac{\begin{vmatrix} x_a - x_c & x - x_c \\ y_a - y_c & y - y_c \end{vmatrix}}{\begin{vmatrix} x_b - x_a & x_c - x_a \\ y_b - y_a & y_c - y_a \end{vmatrix}} , \] \[ \gamma = \frac{\begin{vmatrix} x_b - x_a & x - x_a \\ y_b - y_a & y - y_a \end{vmatrix}}{\begin{vmatrix} x_b - x_a & x_c - x_a \\ y_b - y_a & y_c - y_a \end{vmatrix}} , \]

Let us assume that we are using the RGB model. That is all colors can be obtained by mixing R (red), G (green) and B (blue). Usually, with such a model, the level of each color channel is a number between 0 and 255. To specify the color at any point, we must specify a triplet (R, G, B) where R, G, and B are integers between 0 and 255. They indicate how much of red, green and blue is used in the color. When using Java, there is a built-in class to handle colors. It is called Color. This class has some built-in predefined colors. Here are some examples: Color.black, Color.blue, Color.cyan, Color.gray, Color.green, Color.magenta, Color.orange, Color.pink, Color.red, Color,white, Color.yellow. To get any other color, one uses a statement such as new Color(R,G,B) where R,G,B are integers between 0 and 255.

It is also possible to use a single integer to represent colors. Keeping in mind that an integer has 32 bits, bits 0 − 7 contain the R level, bits 8 − 15 contain the G level and bits 16−23 contain the B level. The remaining bits are unused and set to 0. Let us see now how barycentric coordinates can be used to smoothly color a triangle, given the color of its vertices. Using the notation above, let us assume that C_𝑎 is the color of 𝑎, C_b is the color of b and C_c is the color of c. Each color is in fact a triplet. We will use the notation C_𝑎 = (R_𝑎, G_𝑎, B_𝑎) and similar notation for the remaining points. We would like to color the triangle so that there is a smooth coloring throughout the triangle. We use the fact that by changing the values of α, β, γ between 0 and 1, the point p = α𝑎 + βb + γc will move smoothly inside the triangle. In other words, small changes in α, β, γ will result in small changes in the location of p. We apply this to colors. We let \[ C = \alpha C_a + \beta C_b + \gamma C_c \] (we really do this for every color channel). Small changes in α, β, γ will result in small changes in the color. Therefore, the color will change smoothly as we move within the triangle. To color smoothly a triangle given the color of its vertices, we can use the following algorithm:

1. For each point P = (x, y) inside the triangle, find α, β, γ.
2. Use α, β, γ to interpolate the color of the point from the color of the vertices using relation \[ C = \alpha C_a + \beta C_b + \gamma C_c \]
3. Plot the point with coordinates (x, y) and color computed above.

3D case:   

We use the same notation as in the 2D case. The only difference is that now points have three coordinates. So, we have 𝑎 = (x_𝑎, y_𝑎, z_𝑎), b = (x_b, y_b, z_b) and c = (x_c, y_c, z_c). Barycentric coordinates are extended naturally to 3D triangles and they have the same properties. In other words, we have the same equation for point p = α𝑎 + βb + γc.

The only difference of 3D compared to 2D is that the area of the triangle is alwuas positive independently on orientation. We define the following quantities:

• n is the normal to the triangle T with vertices (𝑎, b, c) in counterclockwise order. In other words, n = (b − 𝑎) × (c − 𝑎).
• n_𝑎 is the normal to T_𝑎, the triangle with area A_𝑎 as shown in figure 1. T_𝑎 = (b, c, p) in counterclockwise order. Thus, n_𝑎 = (c − b) × (p − b).
• n_b is the normal to T_b, the triangle with area A_b as shown in figure 1. T_b = (c, 𝑎, p) in counterclockwise order. Thus, n_b = (𝑎 − c) × (p − c).
• n_c is the normal to T_c, the triangle with area A_c as shown in figure 1. T_c = (𝑎, b, p) in counterclockwise order. Thus, n_c = (b − 𝑎) × (p − 𝑎).

We consider two numerical vectors with entries from ℚ⁴: \[ \mathbf{a} = \left( \frac{3}{7} , \ \frac{2}{5},\ \frac{6}{13} , \ \frac{11}{23} \right) , \quad \mathbf{b} = \left( \frac{7}{15} , \ \frac{5}{12} , \ \frac{23}{47} , \ \frac{5}{11} \right) . \] Their dot product in exact arithmetics is \[ \mathbf{a} \bullet \mathbf{b} = \frac{3481}{4230} \approx 0.8229314420803783 . \]

1. Find the dot product of the following pairs of vectors. \[ {\bf (a)\ \ } \begin{pmatrix} 1 \\ -3 \\ 4 \end{pmatrix} , \quad \begin{pmatrix} 8 \\ 6 \\ 1 \end{pmatrix} \qquad {\bf (b)\ \ } \begin{pmatrix} 3 \\ 5 \\ 6 \end{pmatrix} , \quad \begin{pmatrix} -9 \\ 2 \\ 3 \end{pmatrix} \]
2. how that for any vectors, x, y ∈ ℝⁿ, we have \[ \| \mathbf{x} + \mathbf{y} \|^2 = \| \mathbf{x} \|^2 + 2\,\mathbf{x} \bullet \mathbf{y} + \| \mathbf{y} \|^2 . \] 𝑥 , 𝑦 ∈ 𝑅 𝑛 , we have
3. For vectors u, v ∈ ℝⁿ, show that \( \displaystyle \quad (\mathbf{u} \bullet \mathbf{v}) = \frac{1}{4} \left( \| \mathbf{u} + \mathbf{v} \|^2 - \| \mathbf{u} - \mathbf{v} \|^2 \right) . \)
4. Prove the parallelogram identity: \[ \| \mathbf{u} + \mathbf{v} \|^2 + \| \mathbf{u} - \mathbf{v} \|^2 = 2\, \| \mathbf{u} \|^2 + 2\,\| \mathbf{v} \|^2 . \]
5. What is the angle between the vectors i + j and i + 3j?
6. What is the area of the quadrilateral with vertices at (1, 1), (4, 2), (3, 7) and (2, 3)?
7. Find cos(θ) where θ is the angle between the vectors \[ \left( 3, -2, 7 \right) \quad \mbox{and} \quad \left( 5,3,4 \right) . \]
8. Find cos(θ) where θ is the angle between the vectors \[ \left( \right) \quad \mbox{and} \quad \left( \right) . \]
9. Verify the Cauchy inequality for vectors \[ \left( \right) \quad \mbox{and} \quad \left( \right) . \]
10. Find proj_u(v) where \[ \left( \right) \quad \mbox{and} \quad \left( \right) . \]
11. Find proj_u(v) where \[ \left( \right) \quad \mbox{and} \quad \left( \right) . \]
12. Decompose the vector v into v = v_∥ + v_⊥, where \[ \left( \right) \quad \mbox{and} \quad \left( \right) . \]
13. Show that \[ \mathbf{u} \bullet \left( \mathbf{v} - \mbox{proj}_u (\mathbf{v}) \right) = 0 \] and conclude every vector in &Ropf'ⁿ can be written as the sum of two vectors, one of which is orthogonal and another is parallel to the given vector.
14. Are the vectors u = (2, 3, −1, 4) and v = (?, ?, ?, ?) T orthogonal?

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11. Vector addition