How these 3 vectors form a right triangle?

Question

$ A = 2\hat{i} -2\hat{j} + 3\hat{k}$

$ B = 2\hat{i} -\hat{j} + 3\hat{k}$

$ C = \hat{i} -\hat{j} - \hat{k}$

My textbook demands that these 3 vectors form right triangle. Firstly, I think these 3 vectors don't form a triangle.

Actual Condition for triangle : $ A\pm{B}\pm{C} = 0$ and this is flase for these 3 vectors.

Am I wrong?

Answer 1

Here's what I'd do:

1. Determine which two vectors are perpendicular. You can use the dot product to test this.
2. Verify as you suggested in your question that the three vectors add to zero. You'll need to find the right signs on the vectors. (A triangle has three line segments, but you're given three vectors, each of which has a direction that individually doesn't matter. So take one of the vectors as-is, and try the four possibilities for the signs of the other two when you add them up. You should win on one of them.)

Answer 2

The given vectors are position vectors. To get the vectors representing the sides of the triangle do $A-B, B-C, C-A$. You will find that $\angle B = 90^{\circ}$

Answer 3

Notice that there may be an ambiguity of how you "see" these vectors. If you look at them as if they were three points in $\mathbb{R}^3$, they determine in fact a triangle.

To see if this triangle is right-angled, then you may use Pythagoras Theorem. In this case the distance between $A$ and $B$ is $3$, the distance between $B$ and $C$ is $\sqrt{17}$ and the distance between $C$ and $A$ is $\sqrt{26}$. You can check then that there is a right angle in the point $B$.

Answer 4

$AB=\hat j$, $AC=-\hat i+\hat j-4\hat k$ and $BC=-\hat i-4\hat k$. As $|AB|^2+|BC|^2=1+17=18=|AC|^2$ Pythagoras tells us that the triangle has a right angle at $B$ and $AC = AB + BC$.