How do I find out whether three 3D vectors can form a right angled triangle? 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.
 A: 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 $90^\circ$).
A: The approach in principle is fine: either verify that there's a 90-degree angle (i.e. a zero-dot product) or that the Pythagorean theorem holds. The trouble is that the vectors $\vec{A},\vec{B},\vec{C}$ give the positions of the three vertices but not their relative displacements. So computing $|A|$ as above gives the distance from that vertex to the origin, not the length of one of the sides.
This wouldn't be a problem if one of the vertices was sitting at the origin. But our coordinate system is something we're free to change about by rotations or translations. So we can make that the case if by subtracting one of the vectors from the others, e.g. $$\{\vec{A},\vec{B},\vec{C}\}\mapsto \{0,\vec{B}-\vec{A},\vec{C}-\vec{A}\}.$$ With these in mind, the angle and lengths for the two sides adjoining the point $A$ can be computed by appropriate dot products. This can be repeated for a second vertex to obtain the remaining data. At that point either approach (Pythagoras or angles) will do just fine.
A: First of all, I have to say I haven't heard of using vectors in twelfth grade.  That said, hopefully this will all make sense.
When adding vectors, one normally puts the tail of one vector at the tip of the other and the result is the displacement from the tip of one vector to the tail of the other.  If the triangle were to require all the vectors to be tip to tail to one another, all that would be necessary to check for a triangle would be to sum all 3 vectors.  If you wind back up where you started, the sum would be zero.
This is not the case.  If you relax this requirement, it is also possible that 2 vectors sum to the third.  In this case, $\vec{B}+\vec{C}=\vec{A}$.  So it is possible to draw these vectors such that they form a triangle.
Now for a right triangle, $2$ of the vectors must be orthogonal.  The easiest way to test this is to take the dot product.  Two vectors are orthogonal if their dot product is $0$.  In this case
$$\vec{A}\cdot\vec{C}=3(2)+(-2)(1)+1(-4)=0$$
So vectors $\vec{A}$ and $\vec{C}$ form a right angle.  So they do in fact form a right triangle.
A: Find the dot product of all the pairs of sides of the triangle given. If one of the dot products equal to $0$. 
Applying the formula of $AB\cos\alpha=A\cdot B$. we get $\cos\alpha=0$, from which we get angle to be 90˚. Hence a right angles triangle will be formed.
