Show that the points \(A,B,C\), whose position vectors are respectively \(\vec a=3\hat i-4\hat j-4\hat k\), \(\vec b=2\hat i-\hat j+\hat k\) and \(\vec c=\hat i-3\hat j-5\hat k\), form the vertices of a right-angled triangle.
Show Hint
Find AB, AC, BC and check the Pythagoras relation (or dot product = 0 for two sides).
Step 1: Direct dot-product test:
Instead of comparing squared lengths, test perpendicularity of the two sides meeting at \(A\) directly: \(\overrightarrow{AB}=-\hat i+3\hat j+5\hat k\), \(\overrightarrow{AC}=-2\hat i+\hat j-\hat k\).
Step 2: Computing the dot product:
\(\overrightarrow{AB}\cdot\overrightarrow{AC}=(-1)(-2)+(3)(1)+(5)(-1)=2+3-5=0\).
Step 3: Interpreting the result:
A zero dot product means \(\overrightarrow{AB}\perp\overrightarrow{AC}\), so the angle at vertex \(A\) is exactly \(90^\circ\).
Final Answer:
Since two sides through \(A\) are perpendicular, \(\triangle ABC\) is right-angled at \(A\).\[ \boxed{\text{Right angle at } A} \]