Dot Product of Vectors

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

\[ \vec{a} \cdot \vec{b} = \lVert \vec{a} \rVert \times \lVert \vec{b} \rVert \cos \theta,\]

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

Algebraically, the dot product of two vectors \( \vec{a} = [ a_1 , a_2, \ldots, a_n ] \) and \( \vec{b} = [b_1, b_2, \ldots, b_n ] \) is given by

\[ \vec{a} \cdot \vec{b} = \sum a_i b_i.\]

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

The dot product satisfies the following properties:

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

  2. \(\vec{a} \cdot \vec{a} = \lVert \vec{a} \rVert ^2 \).

  3. Given vectors \( \vec{a} \) and \( \vec{b} \), the projection of \( \vec{a} \) onto \( \vec{b} \) 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 \( OAB\), let \( \vec{OA} = \vec{a}\), \( \vec{OB} = \vec{b}\) and \( \vec{BA} = \vec{c} = \vec{a} - \vec{b} \). Then, we have

\[ \begin{aligned} \lVert\vec{c} \rVert^2 & = \vec{c} \cdot \vec{c} = ( \vec{a} - \vec{b} )( \vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \\ & = \lVert \vec{a} \rVert^2 - 2 \lVert \vec{a} \rVert \lVert \vec{b} \rVert \cos \theta + \lVert \vec{b} \rVert^2. \end{aligned} \]

 

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

\[ \lvert AB \rvert^2 + \lvert CD \rvert^2 - \lvert BC \rvert^2 - \lvert AD \rvert^2 = 2 \vec{AC} \cdot \vec{DB} .\]

Solution: We have

\[ \begin{aligned} \quad \lvert AB \rvert^2 + \lvert CD \rvert^2 - \lvert BC \rvert^2 - \lvert AD \rvert^2 & = ( \vec{b}- \vec{a})^2 + ( \vec{d}- \vec{c})^2 - ( \vec{c}- \vec{b})^2 - ( \vec{d}- \vec{a})^2 \\ & = 2 ( -\vec{b} \cdot \vec{a} - \vec{d} \cdot \vec{c} + \vec{c} \cdot \vec{b} + \vec{d} \cdot \vec{a})\\ & = 2 ( \vec{b} - \vec{d}) \cdot (\vec{c} - \vec{a})\\ & = 2 \vec{ AC} \cdot \vec {DB}. \\ \end{aligned} \]

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

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

\[ \begin{aligned} \sum \lvert A_i P \rvert^2 & = \sum (\vec{P} - \vec{A_i}) \cdot (\vec{P} - \vec{A_i}) \\ & = \sum \left(\vec{P} \cdot \vec{P} + \vec{A} \cdot \vec{A} - 2 \vec{P} \cdot \vec{A_i}\right)\\ & = nr^2 + nr^2 - 2 \vec{P} \cdot \left( \sum \vec{A_i} \right) \\ & = 2nr^2. \\ \end{aligned} \]

Note by Arron Kau
4 years, 1 month 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 - 3 years, 9 months 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 - 3 years, 5 months ago

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good

Vachaspati Mishra - 2 years, 10 months ago

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Nice proof for the last question

Siddharth Iyer - 3 years ago

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