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: Treat the three given vectors as position vectors of points A, B, C:
$A=(2,-1,1)$, $B=(1,-3,-5)$, $C=(3,-4,-4)$.

Step 2: Compute all three pairwise squared distances directly via the distance formula:
$AB^2=(1-2)^2+(-3+1)^2+(-5-1)^2=1+4+36=41$. $BC^2=(3-1)^2+(-4+3)^2+(-4+5)^2=4+1+1=6$. $CA^2=(2-3)^2+(-1+4)^2+(1+4)^2=1+9+25=35$.

Step 3: Test all three possible Pythagorean pairings to find which one balances:
$6+35=41$ ✓ matches $AB^2$; the other two combinations ($41+6=47\ne35$, $41+35=76\ne6$) don't balance.

Final Answer:
Only $BC^2+CA^2=AB^2$ holds, so the right angle is at $C$, confirming a right-angled triangle. \[ \boxed{\text{right angle at }C} \]