Intereting Posts

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This is something that bothered me in my lectures on analytics geometry (because it was given without proof). I can see that $\langle (x,0),(0,y) \rangle=0$ easily, but what about when these two vectors are rotated? I believe I’ve been able to prove it for $2$-dimensions:

$$\langle A,B \rangle=\langle (|A|\cos\theta ,|A| \sin \theta ),(|B|\cos\psi ,|B| \sin \psi)\rangle\\ = |A||B|cos \theta \cos \psi+|A||B|\cos \theta \cos \psi\\=|A||B|(\cos\theta \cos\psi+ \sin \theta \sin \psi)\\=|A||B|\cos (\theta – \psi)$$

Now $\cos( \theta – \psi) =0$ exactly when $ \theta – \psi=n\pi +\cfrac{\pi}{2}, n\in \Bbb{Z}$. That is *basically*, when the difference of the two angles is $\cfrac{\pi}{2},\cfrac{3\pi}{2}$.

- How can I calculate a $4\times 4$ rotation matrix to match a 4d direction vector?
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- which axiom(s) are behind the Pythagorean Theorem
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- draw $\triangle ABC$ in which $AB=5.5$cm, $\angle C =40^{\circ}$ and $BC-AC=2.5$cm
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- Find minimum of $a+b$ under the condition $\frac{m^2}{a^2}+\frac{n^2}{b^2}=1$ where $m,n$ are fixed arguments
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- What hexahedra have faces with areas of exactly 1, 2, 3, 4, 5, and 6 units?
- Line and plane intersection in 3D

Given any nonzero vectors $\vec a, \vec b \in \mathbb R^n$, let $O$ be the origin and let $P$ and $Q$ be the points in $\mathbb R^n$ such that $\vec a = \overrightarrow{OP}$ and $\vec b = \overrightarrow{OQ}$. Let $\vec c = \overrightarrow{PQ}$ so that $\vec a + \vec c = \vec b$.

Observe that $\vec a \perp \vec b$ iff $\Delta OPQ$ satisfies the Pythagorean Theorem, where $PQ$ is the hypotenuse of the candidate right triangle. Indeed, notice that:

\begin{align*}

|\vec a|^2 + |\vec b|^2 = |\vec c|^2

&\iff |\vec a|^2 + |\vec b|^2 = |\vec b – \vec a|^2 \\

&\iff \langle \vec a, \vec a \rangle + \langle \vec b, \vec b \rangle = \langle \vec b + (-\vec a), \vec b + (-\vec a) \rangle \\

&\iff \langle \vec a, \vec a \rangle + \langle \vec b, \vec b \rangle = \langle \vec b, \vec b \rangle + \langle \vec b, -\vec a \rangle + \langle -\vec a, \vec b \rangle + \langle -\vec a, -\vec a \rangle \\

&\iff \langle \vec a, \vec a \rangle + \langle \vec b, \vec b \rangle = \langle \vec b, \vec b \rangle -2\langle \vec a, \vec b \rangle + \langle \vec a, \vec a \rangle \\

&\iff 0 = -2\langle \vec a, \vec b \rangle \\

&\iff \langle \vec a, \vec b \rangle = 0

\end{align*}

- Proofs of Hyperbolic Functions
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