A) right angled triangle
B) isosceles triangle
C) equilateral triangle
D) collinear
Correct Answer: C
Solution :
Let the position vectors \[\overrightarrow{OA}=2\hat{i}+3\hat{j}+4\hat{k},\overrightarrow{OB}=3\hat{i}+4\hat{j}+2\hat{k}\] and \[\overrightarrow{OC}=4\hat{i}+2\hat{j}+3\hat{k}\] Now, \[\overrightarrow{AB}=3\hat{i}+4\hat{j}+2\hat{k}-(2\hat{i}+3\hat{j}+4\hat{k})\] \[=\hat{i}+\hat{j}-2\hat{k}\] \[|\overrightarrow{AB}|=\sqrt{1+1+4}=\sqrt{6}\] \[\overrightarrow{BC}=4\hat{i}+2\hat{j}+3\hat{k}-(3\hat{i}+4\hat{j}+2\hat{k})\] \[=\hat{i}-2\hat{j}+\hat{k}\] \[|\overrightarrow{BC}|=\sqrt{1+4+1}=\sqrt{6}\] \[\overrightarrow{CA}=2\hat{i}+3\hat{j}+4\hat{k}-(4\hat{i}+2\hat{j}+3\hat{k})\] \[=-2\hat{i}+\hat{j}+\hat{k}\] \[|\overrightarrow{CA}|=\sqrt{4+1+1}=\sqrt{6}\] \[\therefore \] \[|\overrightarrow{AB}|=|\overrightarrow{BC}|=\overrightarrow{|CA}|\] \[\Rightarrow \]Given points are the vertices of an equilateral triangle.
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