Vector Algebra Question 345
Question: 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]
Options:
A) Form an isosceles triangle
B) Form a right-angled triangle
C) Are collinear
D) Form a scalene triangle
Show Answer
Answer:
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.
BETA
