A) right angled and isosceles
B) right angled, but not isosceles
C) isosceles but not right angled
D) equilateral
Correct Answer: D
Solution :
Given vertices are \[A(3\hat{i}+\hat{j}+2\hat{k}),\,B(\hat{i}-2\hat{j}+7\hat{k})\]and \[C(-2\hat{i}+3\hat{j}+5\hat{k})\] Now, \[\overrightarrow{AB}=(\hat{i}-2\hat{j}+7\hat{k})-(3\hat{i}+\hat{j}+2\hat{k})\] \[=-2\hat{i}-3\hat{j}+5\hat{k}\] \[\therefore \] \[|\overrightarrow{AB}|=\sqrt{4+9+25}=\sqrt{38}\] \[\overrightarrow{BC}=(-2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}-2\hat{j}+7\hat{k})\] \[=-3\hat{i}+5\hat{j}-2\hat{k}\] \[\therefore \] \[|\overrightarrow{BC}|=\sqrt{38}\] and \[\overrightarrow{CA}=(3\hat{i}+\hat{j}+2\hat{k})-(-2\hat{i}+3\hat{j}+5\hat{k})\] \[=5\hat{i}-2\hat{j}-3\hat{k}\] \[\therefore \] \[|\overrightarrow{CA}|=\sqrt{38}\] Since, \[|\overrightarrow{AB}|=|\overrightarrow{BC}|=|\overrightarrow{CA}|=\sqrt{38}\] \[\therefore \] Hence, triangle is an equilateral triangle.
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