Question

Prove that vectors \(2\hat i-\hat j+\hat k\), \(\hat i-3\hat j-5\hat k\) and \(3\hat i-4\hat j-4\hat k\) are vertices of a right-angled triangle.

Hint:

Compute the three side-vectors, their squared lengths, and check which pair satisfies the Pythagorean relation.

Answer 1

Step 1: Label the points and find the side vectors:
Let \(A=(2,-1,1)\), \(B=(1,-3,-5)\), \(C=(3,-4,-4)\).
\(\vec{AB}=B-A=(-1,-2,-6)\), \(\vec{BC}=C-B=(2,-1,1)\), \(\vec{CA}=A-C=(-1,3,5)\).

Step 2: Compute the squared lengths of the three sides:
\(|AB|^2=1+4+36=41\), \(|BC|^2=4+1+1=6\), \(|CA|^2=1+9+25=35\).

Step 3: Check the Pythagorean relation:
\(|BC|^2+|CA|^2=6+35=41=|AB|^2\). This matches, so the triangle is right-angled with the right angle at the vertex common to sides \(BC\) and \(CA\), which is \(C\).

Final Answer:
Since \(BC^2+CA^2=AB^2\), the triangle is right-angled at \(C\). \[ \boxed{BC^2+CA^2=AB^2=41} \]