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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
3 years, 4 months ago

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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, · 3 years 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. · 2 years, 8 months ago

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good · 2 years, 1 month ago

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Nice proof for the last question · 2 years, 3 months ago

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