May be simple, but how is each term geometrically interpreted in a comprehensive diagram?
We have a Pythagorean relation between dot and cross products here, so how draw it in the plane?
$$|a \times b|^2 + (a.b)^2 = |a |^2 | b|^2. $$
EDIT1:
Trying to depict dot product norm , cross product norm and product of each of two norms in a Pythagorean semi-circle diagram. Cross product is a diagonal of a parallelogram, but dot product is a scalar...
$\endgroup$ 62 Answers
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Given an arbitrary chosen point and vectors $\vec{a}$ and $\vec{b}$.
- Let $A$ be the point with $\vec{OA} = \vec{a}$ and $B$ be the point with $\vec{OB} = \vec{b}$.
- Construct a parallelogram with $O$, $A$, $B$ as vertices. let $C$ be the remaining vertex.
Perpendicular project $A$ and $C$ onto the line $OB$ and let the feet be $A'$ and $C'$ respectively.
We can slide the bottom edge $OB$ of the parallelogram $OBCA$ along the line $OB$ while keeping the top edge $AC$ fixed. When $OB$ reaches the position of $A'C'$, we obtain an rectangle $AA'C'C$ with same area $|\vec{a} \times \vec{b}|$ as parallelogram $OBCA$.
On the plane holding the line $OB$ perpendicular to the axis $AA'$. Rotate $O$ with respect to $A'$ for $90^\circ$ to get the point $O'$. Similarly, rotate $B$ with respect to $C'$ for $90^\circ$ to get the point $B'$. Since $|A'O| = |A'O'|$ and $|AO| = |AO'|$, rectangle $A'O'B'C'$ has area $|\vec{a}\cdot\vec{b}|$ and rectangle $AO'B'C$ has area $|\vec{a}||\vec{b}|$.
For all three rectangles $AA'C'C$, $A'O'B'C'$, $AO'B'C$, one of their side has length $|A'C'| = |OB|$. This implies
$$ \begin{align} |\vec{a} \times \vec{b} | : |\vec{a}\cdot\vec{b}| : |\vec{a}||\vec{b}| & = \verb/Area/(AA'C'C) : \verb/Area/(A'O'B'C') : \verb/Area/(AO'B'C)\\ & = |AA'| : |A'O'| : |AO'| \end{align} $$ Since triangle $AA'O'$ is an right angle triangle, by Pythagorean theorem, we have $$|AA'|^2 + |A'O'|^2 = |AO'|^2 \quad\implies\quad |\vec{a} \times \vec{b} |^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2$$
$\endgroup$ 2 $\begingroup$Taking it easy, wlog let $ \bar b =1$ be a unit vector aligned in direction of $a$. Angle between vectors $a,b$ is $\theta$ and we have for absolute value of cross product
$$ DotPr = a\cos \theta,\, CrossProd = a\sin \theta,\, ScalarProd= a \,$$
seen in Pythagorean setting :
