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For the above diagram, with A [2,1] and N [3,4], and angle AMN = 90 degrees, I would like to confirm if the length of AM is equal to MN. Also, what would be the length of MN.

Here are the two threads where I have raised a similar query but getting apparently different answer for the length of MN and AM. Both the replies confirm though that the length of AM and MN equal.

https://www.canva.com/design/DAE6RUziPE0/ULe8wo10yZfo9tS24aSdWg/view?utm_content=DAE6RUziPE0&utm_campaign=designshare&utm_medium=link&utm_source=sharebutton

https://mathforums.com/threads/relating-dot-product-divided-with-square-of-the-vector-while-changing-basis-of-vector.362851/

If indeed AM and MN are equal, it means that triangle ANM will be right angled isosceles triangle. A geometrical proof will be helpful.

Splendid Digital Solutions
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    Trig hint: the slope of $OA$ is $1/2$ and the slope of $AN$ is $3/1$, then calculate the angle $\angle NAM$. Geometric hint: draw a better diagram and see where $M$ falls, then prove it. – dxiv Mar 07 '22 at 06:32
  • Just to confirm, the length of AN = root of 10. It is incorrect to say that length of MN = root of 10 as mentioned here: https://www.canva.com/design/DAE6RUziPE0/ULe8wo10yZfo9tS24aSdWg/view?utm_content=DAE6RUziPE0&utm_campaign=designshare&utm_medium=link&utm_source=sharebutton – Splendid Digital Solutions Mar 07 '22 at 11:51
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    That's correct. $AN$ is the hypotenuse of a right triangle with sides $1$ and $3$, so $AN=\sqrt{1^2+3^2}$. Also, $AM$ and $MN$ are hypotenuses of right triangles with sides $1$ and $2$, so $AM=MN=\sqrt{1^2+2^2}$. – dxiv Mar 07 '22 at 16:24

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Reformulation of the problem

Euclidean spaces are the adequate setting to discuss elementary geometry. In this case we are considering a two-dimensional euclidean space, i.e. an affine space $E$ with a two-dimensional real translation space $V$ and an inner product on $V$. We assume we are given an orthonormal basis $e$ of $V$, $$V\ni v\mapsto v_e\in\mathbb R^2$$is the induced isomorphism and $O,A,M,N\in E$. We are given that $$\overrightarrow{OA}_e=\begin{pmatrix}2\\1\end{pmatrix}$$and$$\overrightarrow{ON}_e=\begin{pmatrix}3\\4\end{pmatrix}$$ In addition, $\overrightarrow{ON}=\overrightarrow{OM}+\overrightarrow{MN}$ is the decomposition of $\overrightarrow{ON}$ as the sum of a vector parallel to $\overrightarrow{OA}$ and a vector perpendicular to $\overrightarrow{OA}$. All we need to do to solve your problem is to determine the vector $\overrightarrow{OM}$: If we know $\overrightarrow{OM}$, then the distance from $M$ to $N$ is given by $$\|\overrightarrow{MN}\|=\|\overrightarrow{ON}-\overrightarrow{OM}\|$$ and the distance from $A$ to $M$ is given by $$\|\overrightarrow{AM}\|=\|\overrightarrow{OM}-\overrightarrow{OA}\|$$ So now our problem consists in determining $\overrightarrow{OM}$.

Solution of the reformulated problem

Suppose $u,v\in V$, $u\neq 0$ and we want to determine the decomposition $v=v_\parallel+v_\perp$ where $v_\parallel$ is parallel to $u$ and $v_\perp$ is perpendicular to $u$. It is well known that $$v_\parallel=\frac{u\cdot v}{u\cdot u}u$$ and this is not difficult to proof, but let me give an intuitive explanation: Using $\|u\|=\sqrt{u\cdot u}$ and $u\cdot v=\|u\|\|v\|\cos\theta$ we obtain $$v_\parallel=\frac{u\cdot v}{\|u\|}\frac{u}{\|u\|}=\|v\|\cos\theta\frac{u}{\|u\|}$$ i.e. $v_\parallel$ is the vector with the length $\|v\|\cos\theta$ pointing in the same direction as $u$ (if $\cos\theta>0$).

If we apply this to our problem, we obtain $$\overrightarrow{OM}=\frac{\overrightarrow{OA}\cdot\overrightarrow{ON}}{\overrightarrow{OA}\cdot\overrightarrow{OA}}\overrightarrow{OA}$$In addition, since $e$ is orthonormal, we have that $v\cdot w=v_e\cdot w_e$ for all $v,w\in V$. So now all you need to do is plug in the numbers and you'll see that the answer to your question is YES.

Filippo
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