Question

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.

Hint:

Find AB, AC, BC and check the Pythagoras relation (or dot product = 0 for two sides).

Answer 1

Step 1: Finding the side vectors:
\(\overrightarrow{AB}=\vec b-\vec a=(2-3)\hat i+(-1+4)\hat j+(1+4)\hat k=-\hat i+3\hat j+5\hat k\). \(\overrightarrow{AC}=\vec c-\vec a=(1-3)\hat i+(-3+4)\hat j+(-5+4)\hat k=-2\hat i+\hat j-\hat k\).

Step 2: Finding the third side:
\(\overrightarrow{BC}=\vec c-\vec b=(1-2)\hat i+(-3+1)\hat j+(-5-1)\hat k=-\hat i-2\hat j-6\hat k\).

Step 3: Computing squared lengths:
\(|\overrightarrow{AB}|^2=1+9+25=35\), \(|\overrightarrow{AC}|^2=4+1+1=6\), \(|\overrightarrow{BC}|^2=1+4+36=41\).

Step 4: Applying the Pythagoras test:
\(|\overrightarrow{AB}|^2+|\overrightarrow{AC}|^2=35+6=41=|\overrightarrow{BC}|^2\), so the triangle is right-angled at \(A\).

Final Answer:
\[ \boxed{\triangle ABC \text{ is right-angled at } A} \]