Dot Product of Vectors

The dot product of two vectors a \vec{a} and b \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
5 years, 4 months ago

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Comments

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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?

Thanks,

John Muradeli - 5 years, 1 month ago

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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 - 4 years, 9 months ago

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good

Vachaspati Mishra - 4 years, 2 months ago

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Nice proof for the last question

Siddharth Iyer - 4 years, 4 months ago

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