[Math] How to find out whether three 3D vectors can form a right angled triangle

geometry

I am asking this question for my son who is about finish the twelfth grade.

I have already seen this question, however that did not actually answer my query.

I have three vectors,

\begin{align*}
\vec{A} &= 3\hat{i} – 2\hat{j} + \hat{k}\\
\vec{B} &= \hat{i} – 3\hat{j} + 5\hat{k}\\
\vec{C} &= 2\hat{i} + \hat{j} – 4\hat{k}
\end{align*}

and I need to find out whether they can form a right angled triangle.

One way to attack the problem will be to find out the length
of the vectors.

$|A|^2 = 9+4+1 = 14$

$|B|^2 = 1+9+25 = 35$

$|C|^2 = 4+1+16 = 21$

And then apply Pythagoras theorem,

$|B|^2 = |A|^2 + |C|^2 = 35$ .

Also, we need to check whether the angle between these $\vec{A}$ and
$\vec{C}$ is a right angle,

$\vec{A}\cdot\vec{C} = 6-2-4 = 0$.

Now, what is wrong if I do not use Pythagoras and find out the
three angles between the pairs of vectors and then simply check that
one is a right angle and the sum of the three angles is $180^\circ$?
Definitely at the same time I will need to check that the sum of two sides
is larger than the largest side.

Best Answer

I think question is "whether they can form a right angled triangle" without rotation, so $2$ conditions only are to check:

  • vectors can be placed "in one plane": one of $\vec{A}\pm \vec{B}\pm \vec{C}$ must be $0$;
  • $2$ of them form $90^\circ$: dot product of some $2$ vectors is $0$.

If these $2$ conditions are true, then Pythagoras theorem is excessive checking.

Here we have:
$\vec{A}-\vec{B}-\vec{C}=0$,
$\vec{A}\cdot\vec{C}=0$.

That say that vectors $\vec{A},\vec{B},\vec{C}$ form right triangle.

Example without $1$st condition:
$\vec{A}=(3,0,0),\quad\vec{B}=(0,4,0),\quad\vec{C}=(0,0,5)$. (don't lie in one plane).

Example without $2$nd condition:
$\vec{A}=(3,3,3),\quad\vec{B}=(1,2,1),\quad\vec{C}=(2,1,2)$. (there are no angle $90^\circ$).

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