Dot Product

Dot Product #

Revised

10 Mar 2023

Dot Product #

Algebraic Definition #

To find the dot product of two vectors, we multiply corresponding components and add. The result of the dot product is not a vector, but a number called a scalar.

If $a = ⟨ a_{1}, . . ., a_{n} ⟩$ and $b = ⟨ b_{1}, . . ., b_{n} ⟩$ , then the dot product of $a$ and $b$ is the number $a \cdot b$ given by

$\begin{array}{r} a \cdot b = \sum_{i = 1}^{n} a_{i} b_{i} = a_{1} b_{1} + . . . + a_{n} b_{n} \end{array}$

If $a$ and $b$ are interpreted as column vectors, the dot product can be written as a matrix product.

$\begin{array}{r} a \cdot b = a^{⊤} b = [\begin{array}{c} a_{1} & \dots & a_{n} \end{array}] [\begin{array}{c} b_{1} \\ ⋮ \\ b_{n} \end{array}] \end{array}$

Geometric Definition #

Let $θ$ be the angle between the representations of $a$ and $b$ that start at the origin, where $0 \leq θ \leq π$ .

In other words, let $a$ and $b$ be the line segments $\vec{O A}$ and $\vec{O B}$ , respectively.

$a \cdot b = ‖ a ‖ ‖ b ‖ \cos θ$

If $a$ and $b$ are parallel, then $θ = 0$ .

If $a$ and $b$ are antiparallel, then $θ = π$ .

Proof

from the Law of Cosines

$\begin{aligned} | A B |^{2} & = | O A |^{2} + | O B |^{2} - 2 | O A | | O B | \cos θ \\ ⟹ \\ | a - b |^{2} & = | a |^{2} + | b |^{2} - 2 | a | | b | \cos θ \end{aligned}$

$\begin{aligned} | a - b |^{2} & = (a - b) \cdot (a - b) \\ = a \cdot (a - b) - b \cdot (a - b) \\ = a \cdot a - a \cdot b - b \cdot a + b \cdot b \\ = | a |^{2} - 2 a \cdot b + | b |^{2} \end{aligned}$

$\begin{aligned} | a - b |^{2} & = | a |^{2} + | b |^{2} - 2 | a | | b | \cos θ \\ ⟹ \\ | a |^{2} - 2 a \cdot b + | b |^{2} & = | a |^{2} + | b |^{2} - 2 | a | | b | \cos θ \\ ⟹ \\ - 2 a \cdot b & = - 2 | a | | b | \cos θ \\ ⟹ \\ a \cdot b & = | a | | b | \cos θ \end{aligned}$

$◼$

Corollary: the angle between two vectors #

If $θ$ is the angle between the nonzero vectors $a$ and $b$ , then

$\begin{array}{r} \cos θ = \frac{a \cdot b}{| a | | b |} \end{array}$

Properties of the Dot Product #

Dot Product of a vector with itself #

Let $a \in R^{n}$ be a vector.

$a \cdot a = | a |^{2}$

Proof

$\begin{aligned} a \cdot a & = ⟨ a_{1}, . . ., a_{n} ⟩ \cdot ⟨ a_{1}, . . ., a_{n} ⟩ \\ = a_{1} a_{1} + . . . + a_{n} a_{n} \\ = a_{1}^{2} + . . . + a_{n}^{2} \\ = {(\sqrt{a_{1}^{2} + . . . + a_{n}^{2}})}^{2} \\ = | a |^{2} \end{aligned}$

$◼$

Commutativity #

Let $a, b \in R^{n}$ be vectors.

$a \cdot b = b \cdot a$

Proof

$\begin{aligned} a \cdot b & = ⟨ a_{1}, . . ., a_{n} ⟩ \cdot ⟨ b_{1}, . . ., b_{n} ⟩ \\ = a_{1} b_{1} + . . . + a_{n} b_{n} \\ = b_{1} a_{1} + . . . + b_{n} a_{n} \\ = ⟨ b_{1}, . . ., b_{n} ⟩ \cdot ⟨ a_{1}, . . ., a_{n} ⟩ \\ = b \cdot a \end{aligned}$

$◼$

Distributivity over vector addition #

Let $a, b, c \in R^{n}$ be vectors.

$a \cdot (b + c) = a \cdot b + a \cdot c$

Proof

$\begin{aligned} a \cdot (b + c) & = ⟨ a_{1}, . . ., a_{n} ⟩ \cdot (⟨ b_{1}, . . ., b_{n} ⟩ + ⟨ c_{1}, . . ., c_{n} ⟩) \\ = ⟨ a_{1}, . . ., a_{n} ⟩ \cdot ⟨ b_{1} + c_{1}, . . ., b_{n} + c_{n} ⟩ \\ = a_{1} (b_{1} + c_{1}) + . . . + a_{n} (b_{n} + c_{n}) \\ = a_{1} b_{1} + a_{1} c_{1} + . . . + a_{n} b_{n} + a_{n} c_{n} \\ = (a_{1} b_{1} + . . . + a_{n} b_{n}) + (a_{1} c_{1} + . . . + a_{n} c_{n}) \\ = ⟨ a_{1}, . . ., a_{n} ⟩ \cdot ⟨ b_{1}, . . ., b_{n} ⟩ + ⟨ a_{1}, . . ., a_{n} ⟩ \cdot ⟨ c_{1}, . . ., c_{n} ⟩ \\ = a \cdot b + a \cdot c \end{aligned}$

$◼$

Scalar Multiplication #

Let $a, b \in R^{n}$ be vectors and $c \in R$ a scalar.

$(c a) \cdot b = c (a \cdot b) = a \cdot (c b)$

Proof

$\begin{aligned} (c a) \cdot b & = (c ⟨ a_{1}, . . ., a_{n} ⟩) \cdot ⟨ b_{1}, . . ., b_{n} ⟩ \\ = ⟨ c a_{1}, . . ., c a_{n} ⟩ \cdot ⟨ b_{1}, . . ., b_{n} ⟩ \\ = c a_{1} b_{1} + . . . + c a_{n} b_{n} \\ = c (a_{1} b_{1} + . . . + a_{n} b_{n}) \\ = c (⟨ a_{1}, . . ., a_{n} ⟩ \cdot ⟨ b_{1}, . . ., b_{n} ⟩) \\ = c (a \cdot b) \\ = a_{1} c b_{1} + . . . + a_{n} c b_{n} \\ = ⟨ a_{1}, . . ., a_{n} ⟩ \cdot ⟨ c b_{1}, . . ., c b_{n} ⟩ \\ = ⟨ a_{1}, . . ., a_{n} ⟩ \cdot (c ⟨ b_{1}, . . ., b_{n} ⟩) \\ = a \cdot (c b) \end{aligned}$

$◼$

Dot Product with the zero vector #

Let $a \in R^{n}$ be a vector.

$0 \cdot a = 0$

Proof

$\begin{aligned} 0 \cdot a & = ⟨ 0, . . ., 0 ⟩ \cdot ⟨ a_{1}, . . ., a_{n} ⟩ \\ = 0 \cdot a_{1} + . . . + 0 \cdot a_{n} \\ = 0 \end{aligned}$

$◼$

Exercises #

$⟨ 2, 4 ⟩ \cdot ⟨ 3, - 1 ⟩ = 2 \cdot 3 + 4 \cdot - 1 = 6 - 4 = 2$

$⟨ - 1, 7, 4 ⟩ \cdot ⟨ 6, 2, - \frac{1}{2} ⟩ = - 1 \cdot 6 + 7 \cdot 2 + 4 \cdot - \frac{1}{2} = - 6 + 14 - 2 = 6$

$(i + 2 j - 3 k) \cdot (2 j - k) = 1 \cdot 0 + 2 \cdot 2 + - 3 \cdot - 1 = 4 + 3 = 7$

Terms #

[W] Dot Product

Dot Product

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