question_answer

The position vectors of the points A, B, C are \[(2\mathbf{i}+\mathbf{j}-\mathbf{k}),\] \[(3\mathbf{i}-2\mathbf{j}+\mathbf{k})\] and \[(\mathbf{i}+4\mathbf{j}-3\mathbf{k})\] respectively. These points                                                                         [Kurukshetra CEE 2002]

A) Form an isosceles triangle

B) Form a right-angled triangle

C) Are collinear

D) Form a scalene triangle

Correct Answer: C

Solution :

\[\overrightarrow{AB}=(3-2)\mathbf{i}+(-2-1)\mathbf{j}+(1+1)\mathbf{k}=\mathbf{i}-3\mathbf{j}+2\mathbf{k}\] \[\overrightarrow{BC}=(1-3)\mathbf{i}+(4+2)\mathbf{j}+(-3-1)\mathbf{k}=-2\mathbf{i}+6\mathbf{j}-4\mathbf{k}\] \[\overrightarrow{CA}=(2-1)\mathbf{i}+(1-4)\mathbf{j}+(-1+3)\mathbf{k}=\mathbf{i}-3\mathbf{j}-2\mathbf{k}\] \[|\overrightarrow{AB}|\,=\sqrt{1+9+4}=\sqrt{14}\] \[|\overrightarrow{BC}|\,=\sqrt{4+36+16}=\sqrt{56}=2\sqrt{14}\] \[|\overrightarrow{CA}|\,=\sqrt{1+9+4}=\sqrt{14}\] So, \[|\overrightarrow{AB}|+|\overrightarrow{AC}|=\,|\,\overrightarrow{BC}\,|\] and angle between AB and BC is 180°.  Points A, B, C cannot form an isosceles triangle. Hence A, B, C are collinear.