Can one prove that whether a triangle is a right-angled triangle just by looking at the position vectors of its vertices

Question

The position vectors of the vertices of the triangle are given as: \begin{align*} \overrightarrow{a} &= 3\hat{i}-4\hat{j} -4\hat{k} \\ \overrightarrow{b} &= 2\hat{i}-\hat{j} +\hat{k} \\ \overrightarrow{c} &= \hat{i}-3\hat{j} - 5\hat{k} \end{align*} One can simply find out the displacement vectors and then find the square of magnitudes and use Pythagoras theorem, in this specific case, it does work out that $\overrightarrow{b}\cdot\overrightarrow{c}=0$, but I wanted to know that if it's a general result that whether we can simply predict if the triangle will be right-angled just by looking at the position vectors of the vertices.

Answer 1

Example of a non-right-angled triangle where all dot products of position vectors vanish: \begin{align} \vec a &= \hat i\\ \vec b &= \hat j\\ \vec c &= \hat k \end{align}

Example of a right-angled triangle where no products of position vectors are zero: \begin{align} \vec a &= \hat i\\ \vec b &= \hat i + \hat j\\ \vec c &= \hat i + \hat k \end{align}

Thus orthogonality of position vectors is neither necessary nor sufficient for a triangle to be right-angled.

Answer 2

You can compute the dot products of all three pairs of sides, so $(a-b)\cdot (a-c)$ and similar. If any one is $0$, the two vectors involved are perpendicular and the triangle is right. Similarly if none are zero, the triangle is not right.

Added: it is not necessary that the dot product the vectors from the origin to two points is zero. That only happens when the angle from one point to the origin to the other is right, which is independent of the triangle being right.