Given points are A(-3/2,3), B(6,-2) and C(-3,4)
\(\vec{AB}\) = \(\vec{OB}\) - \(\vec{OA}\)
= (6\(\hat i\) - 2\(\hat j\)) - (-3/2\(\hat i\) + 3\(\hat j\))
= (6 + 3/2)\(\hat i\) - (2+3)\(\hat j\)
= 15/2 \(\hat i\) - 5\(\hat j\)
\(\vec{BC}\) = \(\vec{OC}\) - \(\vec{OB}\)
(-3\(\hat i\) + 4\(\hat j\)) - (6\(\hat i\) - 2\(\hat j\))
= (-3 - 6)\(\hat i\) + (4 + 2)\(\hat j\)
= - 9\(\hat i\) + 6\(\hat j\)
Now, \(\vec{AB}\) . \(\vec{BC}\) = (15/2\(\hat i\) - 5\(\hat j\)) . (-9\(\hat i\)+ 6\(\hat j\))
= 15/2 x -9 - 5 x 6
= \(\cfrac{-135}2\) - 30
= \(\cfrac{-135-60}2\) = \(\cfrac{-195}2\)
\(\left|\vec{AB}\right|\) = \(\sqrt{(\cfrac{15}2)^2+(-5)^2}\)
= \(\sqrt{\cfrac{225}4+25}\)
= \(\sqrt{\cfrac{{225}+{100}}4}\) = \(\sqrt{\cfrac{325}2}\)
= \(\cfrac{5\sqrt{13}}2\)
\(\left|\vec{BC}\right|\) = \(\sqrt{(-9)^2+6^2}\)
= \(\sqrt{81+36}\) = \(\sqrt{117}\)
= \(3\sqrt{13}\)
Cos θ = \(\cfrac{{\vec{AB}}.{\vec{BC}}}{{|\vec{AB}|}|{\vec{BC}|}}\)
= \(\cfrac{\cfrac{-195}2}{\cfrac{5\sqrt{13}}{2}\times3\sqrt{13}}\) = - 1 = cos \(\pi\)
Hence, angle between \(\vec {AB}\) and \(\vec {BC}\) is 180°
\(\therefore\) Points A,B and C are collinear
