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Wize University Linear Algebra Textbook > Products of Vectors

Dot Product

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Dot Product

Multiplying vectors by scalars is simple.
Multiplying vectors by vectors is not as straightforward, and there are two kinds of vectors products:
  1. Dot product: takes two vectors in Rn\reals^n and produces a single scalar
  2. Cross product: takes two vectors in R3\reals^3 and produces another vector in R3\reals^3

The dot product (or scalar product) of u⃗=[u1u2⋮un]\vec{u}= \begin{bmatrix} u_1\\ u_2\\ \vdots\\u_n\\ \end{bmatrix}and v⃗=[v1v2⋮vn]\vec{v}= \begin{bmatrix} v_1\\ v_2\\ \vdots\\v_n\\ \end{bmatrix} is defined to be:
u⃗⋅v⃗ = u1v1 + u2v2 + ⋯ + unvn  ∈R\boxed{\quad \vec{u} \cdot \vec{v} \ =\ u_1v_1\ +\ u_2v_2 \ +\ ⋯\ +\ u_n v_n \quad} \ \ \in \reals

Watch Out!
Like vector addition, the dot product only works if the vectors are in the same space!


Example
Given u⃗=⟨4,−2,3⟩\vec{u}=\lang 4,-2,3 \rang and v⃗=⟨5,4,−1⟩\vec{v}=\lang 5,4,-1 \rang, compute u⃗⋅v⃗\vec u \cdot \vec v.
u⃗⋅v⃗=[4−23]⋅[54−1]=4⋅5 + (−2)⋅4 + 3⋅(−1)\vec u \cdot \vec v = \begin{bmatrix} \colorOne{4}\\ \colorTwo{-2}\\ \colorThree{3} \end{bmatrix} \cdot \begin{bmatrix} \colorOne{5}\\ \colorTwo{4}\\ \colorThree{-1} \end{bmatrix} = \colorOne{4}\cdot\colorOne{5} \ +\ (\colorTwo{-2})\cdot\colorTwo{4} \ +\ \colorThree{3}\cdot(\colorThree{-1})
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Example: Dot Product

Part A)

Given u⃗=⟨1,4,−2⟩\vec{u}=\lang 1,4,-2\rang and v⃗=⟨0,−2,5⟩\vec{v}=\lang 0,-2,5\rang, find u⃗⋅v⃗\vec u \cdot \vec{v}.

u⃗⋅v⃗=(1)(0)+(4)(−2)+(−2)(5)=−8−10=−18\begin{aligned} \vec u \cdot \vec v &=(1)(0)+(4)(−2)+(−2)(5)\\ &= -8-10\\ &= \boxed{-18} \end{aligned}

Part B)

If [2−13k]⋅[k91]=5\begin{bmatrix} 2\\ -\frac{1}{3}\\ k\\ \end{bmatrix} \cdot \begin{bmatrix} k\\ 9\\ 1\\ \end{bmatrix} =5, find the value of kk.

(2)(k)+(−13)(9)+(k)(1)=52k−3+k=53k=5+3k=83\begin{aligned} \left(2\right)\left(k\right)+\left(-\frac{1}{3}\right)\left(9\right)+\left(k\right)\left(1\right) &= 5\\ 2k - 3 + k &= 5\\ 3k&= 5+3\\ k&=\boxed{\dfrac{8}{3}} \end{aligned}

Practice: Dot Product

Fill in the blanks.

Part A)

[−8−243]⋅[144−1]=\begin{bmatrix} -8\\ -2\\ 4\\ 3 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 4\\ 4\\ -1 \end{bmatrix} =

Part B)

Given u⃗=[−23]\vec u= \begin{bmatrix} -2\\ 3\\ \end{bmatrix} and v⃗=[5d]\vec v= \begin{bmatrix} 5\\ d\\ \end{bmatrix}, and u⃗⋅v⃗=20\vec u \cdot \vec v = 20,
then d =
.

Extra Practice
133 - FML 3 - 18.1W e.g. 20
For two vectors v⃗\bcb{\vv} and w⃗\bcb{\vw} ∈  I ⁣Rn\bcb{\in}\; \Rn{n}, v⃗⋅w⃗:=‾\bcb{\vv \cdot \vw :=\underline{\qquad\qquad\qquad}}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 40_$\tkcth{Mock F}\tkct{?}$_$\tkct{no soln / vid}$
If u⃗=<1,−2,3>\bcb{\vec{u} = \left< 1, -2, 3 \right>} and v⃗=<2, −1, 3>\bcb{\vec{v} = \left< 2,\, -1,\, 3 \right>}, find (2u⃗)⋅(3v⃗)\bcb{(2\vec{u})\cdot (3\vec{v})}.
Example: Orthogonal Vectors
Example:
Find all values of kk such that the vectorsv⃗=(k, 1)\vec{v}=\left(k,\ 1\right) and u⃗=(k,−1)\vec{u}=\left(k,-1\right) are orthogonal.
Practice Question: Orthogonal Vectors
Practice Question: Orthogonal Vectors
Find the value(s) of cc such that u⃗=(−1,c,2) \vec{u}=(-1,c,2)\ and v⃗=(−7, c, −4)\vec{v}=\left(-7,\ c,\ -4\right) are orthogonal.
Practice Question: Orthogonal Vectors
Practice Question: Orthogonal Vectors
Find the value(s) of cc such that u⃗=(−1,c,2) \vec{u}=(-1,c,2)\ and v⃗=(−7, c, −4)\vec{v}=\left(-7,\ c,\ -4\right) are orthogonal.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 52
If ∣u⃗∣=2\bcb{|{ \vec{u}| } = 2} and ∣v⃗∣=3\bcb{|{ \vec{v} |} = 3} and u⃗\bcb{\vec{u}} has the same direction as v⃗\bcb{\vec{v}} then (3u⃗)⋅(2v⃗)\bcb{(3\vec{u}) \cdot (2\vec{v})} is:
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | $\tkco{duplicate to mock ✓}$ 133 - FML 3 - 18.1W e.g. 46_$\tkcth{Mock F}\tkct{?}$_$\tkct{vid soln only}$
If ∣u⃗∣=2\bcb{|{\vec{u}|} = 2} and ∣v⃗∣=3\bcb{|{ \vec{v} |} = 3}, where u⃗⋅v⃗=2\bcb{\vec{u} \cdot \vec{v} = 2}, find ∣u⃗−v⃗∣\bcb{|{\vec{u} - \vec{v} }|}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 54
Find the value of k\bcb{k} such that the vector u⃗=<1,k,3>\bcb{\vec{u} = \left< 1, k, 3 \right>} is perpendicular to the vector v⃗=<2,−1,k>\bcb{\vec{v} = \left< 2, -1, k\right>}.
If 𝑢⃗=(2,0,−3),𝑣⃗=(−2,2,−4),𝑢⃗ = (2, 0, −3), \vec𝑣 = (−2, 2, −4), and 𝑤⃗=(1,2,1)𝑤⃗= (1,2,1), compute u⃗ ∙ v⃗\vec{u}\ \bullet\ \vec{v}.
Practice Question: Vector Product
Practice Question: Vector Product
Given that 𝑢⃗=(1,1,−1)𝑢⃗=(1,1,−1) and 𝑣⃗=(0,−2,2),𝑣⃗=(0,−2,2), compute (𝑢⃗⋅𝑣⃗)(𝑣⃗×𝑢⃗).(𝑢⃗\cdot𝑣⃗)(𝑣⃗\times𝑢⃗).
Find the value(s) of kk(if any) such that the vectors u→=(1,k,1,0,−1)\overrightarrow{u}=\left(1,k,1,0,-1\right)and v→=(k,2,0,3,−4)\overrightarrow{v}=\left(k,2,0,3,-4\right)are orthogonal.
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ | Basic dot product.
The dot product v⃗⋅w⃗\vv \cdot \vw, for v⃗=<−1, 1, −1>\vv = \rowvecth{-1}{1}{-1} and w⃗=<−2, −2, 3>\vw = \rowvecth{-2}{-2}{3}, is equal to:
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ |
A unit vector perpendicular to the vector w⃗=<2, 5>\vw = \rowvec{2}{5} is:
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ |
A vector perpendicular to the vector connecting the points P1(2,2,2)P_1(2,2,2) and P2(1,−1,1)P_2(1,-1,1) is:
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ |
A vector perpendicular to the vector connecting the points P1(0,1,−2)P_1(0,1,-2) and P2(−2,0,1)P_2(-2,0,1) is:
Dot and Cross Products: Vector Norm
If u⃗, v⃗, w⃗  ∈ R3\vec{u},\ \vec{v},\ \vec{w}\ \ \in\ R^3, which of the following operations is/are defined?
(i.e. which of the following can be calculated?)
Select all that are correct.
Unit Vectors: Dot Product
Consider the vectors u=i−4k\bm{u}=\bm{i}-4\bm{k} and v=3i−2j+k\bm{v}=3\bm{i}-2\bm{j}+\bm{k} in R3\reals^3. Find (if possible) 3v⋅(k−2u)3\bm{v}\cdot(\bm{k}-2\bm{u}).