Position vectors $2i-j+k$, $i-3j-5k$ and $3i-4j-4k$ form a triangle with sides $\sqrt{41}$, $\sqrt{35}$ and $\sqrt{6}$. This shows the triangle is:

1. Compute the three side lengths from the position vectors: $|AB|=\sqrt{41}$, $|AC|=\sqrt{35}$, $|BC|=\sqrt{6}$. 2. Pythagoras' theorem holds when the square of the longest side equals the sum of the squares of the other two. 3. The longest side is $\sqrt{41}$, so test $|AC|^2+|BC|^2=35+6=41=|AB|^2$. 4. The relation holds exactly, so the angle opposite $AB$ is a right angle. 5. Equivalently $\vec{AC}\cdot\vec{BC}=0$, which confirms the same result. Writing $41+6=35$ pairs the wrong sides, since the longest side must stand alone. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 3: Vectors_