Dot Product of Vectors

The dot product of two vectors \( \vec{a} \) and \( \vec{b} \) is

ab=a×bcosθ, \vec{a} \cdot \vec{b} = \lVert \vec{a} \rVert \times \lVert \vec{b} \rVert \cos \theta,

where θ \theta is the angle between the vectors a \vec{a} and b \vec{b} .

Algebraically, the dot product of two vectors a=[a1,a2,,an] \vec{a} = [ a_1 , a_2, \ldots, a_n ] and b=[b1,b2,,bn] \vec{b} = [b_1, b_2, \ldots, b_n ] is given by

ab=aibi. \vec{a} \cdot \vec{b} = \sum a_i b_i.

The equivalence of these two forms can be shown through the cosine rule.

The dot product satisfies the following properties:

  1. If ab=0 \vec{a} \cdot \vec{b} = 0, then θ=90 \theta = 90^\circ, implying a \vec{a} and b \vec{b} are orthogonal.

  2. aa=a2\vec{a} \cdot \vec{a} = \lVert \vec{a} \rVert ^2 .

  3. Given vectors a \vec{a} and b \vec{b} , the projection of a \vec{a} onto b \vec{b} is given by \vec{a}_\vec{b} = \frac { \vec{a} \cdot \vec{b} } { \lVert \vec{b} \rVert} . This formula is useful for finding the (perpendicular) distance of a point to a line (or hyperplane).

Worked Examples

1. Derive the law of cosines.

Solution: In triangle OAB OAB, let OA=a \vec{OA} = \vec{a}, OB=b \vec{OB} = \vec{b} and BA=c=ab \vec{BA} = \vec{c} = \vec{a} - \vec{b} . Then, we have

c2=cc=(ab)(ab)=aa2ab+bb=a22abcosθ+b2. \begin{aligned} \lVert\vec{c} \rVert^2 & = \vec{c} \cdot \vec{c} = ( \vec{a} - \vec{b} )( \vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \\ & = \lVert \vec{a} \rVert^2 - 2 \lVert \vec{a} \rVert \lVert \vec{b} \rVert \cos \theta + \lVert \vec{b} \rVert^2. \end{aligned}


2. Consider four points in space A,B,C,D A, B, C, D. Show

AB2+CD2BC2AD2=2ACDB. \lvert AB \rvert^2 + \lvert CD \rvert^2 - \lvert BC \rvert^2 - \lvert AD \rvert^2 = 2 \vec{AC} \cdot \vec{DB} .

Solution: We have

AB2+CD2BC2AD2=(ba)2+(dc)2(cb)2(da)2=2(badc+cb+da)=2(bd)(ca)=2ACDB. \begin{aligned} \quad \lvert AB \rvert^2 + \lvert CD \rvert^2 - \lvert BC \rvert^2 - \lvert AD \rvert^2 & = ( \vec{b}- \vec{a})^2 + ( \vec{d}- \vec{c})^2 - ( \vec{c}- \vec{b})^2 - ( \vec{d}- \vec{a})^2 \\ & = 2 ( -\vec{b} \cdot \vec{a} - \vec{d} \cdot \vec{c} + \vec{c} \cdot \vec{b} + \vec{d} \cdot \vec{a})\\ & = 2 ( \vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a})\\ & = 2 \vec{ AC} \cdot \vec {DB}. \\ \end{aligned}

3. In a circle Γ \Gamma with center O O and radius r r, we have inscribed a regular n n-gon A1A2An A_1 A_2 \ldots A_n. If P P is any point on the circumference, show that AiP2=2nr2 \sum | A_i P |^2 = 2nr^2 .

Solution: Let O O be the origin. Then A1++An=0 \vec{A_1} + \ldots + \vec{A_n} = 0. Hence,

AiP2=(PAi)(PAi)=(PP+AA2PAi)=nr2+nr22P(Ai)=2nr2. \begin{aligned} \sum \lvert A_i P \rvert^2 & = \sum (\vec{P} - \vec{A_i}) \cdot (\vec{P} - \vec{A_i}) \\ & = \sum \left(\vec{P} \cdot \vec{P} + \vec{A} \cdot \vec{A} - 2 \vec{P} \cdot \vec{A_i}\right)\\ & = nr^2 + nr^2 - 2 \vec{P} \cdot \left( \sum \vec{A_i} \right) \\ & = 2nr^2. \\ \end{aligned}

Note by Arron Kau
6 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

I don't get that last one, #3: Why is Sum of A vectors up to An zero? Also, what is the meaning of A1A2...An, the product of all -gons before the n gon and including?


John M. - 6 years, 1 month ago

Log in to reply

the sum of A VECTORS adds up to zero as it results into a closed structure and hence the addition resulting in a zero resultant vector.

Vivek Rao - 5 years, 9 months ago

Log in to reply


Vachaspati Mishra - 5 years, 2 months ago

Log in to reply

Nice proof for the last question

Siddharth Iyer - 5 years, 4 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...