Vector Algebra Question 279
Question: The point having position vectors $ 2\mathbf{i}+3\mathbf{j}+4\mathbf{k}, $ $ 3\mathbf{i}+4\mathbf{j}+2\mathbf{k}, $ $ 4\mathbf{i}+2\mathbf{j}+3\mathbf{k} $ are the vertices of
[EAMCET 1988]
Options:
A) Right angled triangle
B) Isosceles triangle
C) Equilateral triangle
D) Collinear
Show Answer
Answer:
Correct Answer: C
Solution:
- Here, $ \overrightarrow{OA}=2\mathbf{i}+3\mathbf{j}+4\mathbf{k}, $ $ \overrightarrow{OB}=3\mathbf{i}+4\mathbf{j}+2\mathbf{k} $ $ \overrightarrow{OC}=4\mathbf{i}+2\mathbf{j}+3\mathbf{k} $ So, $ \overrightarrow{AB}=\mathbf{i}+\mathbf{j}-2\mathbf{k}, $ $ \overrightarrow{BC}=\mathbf{i}-2\mathbf{j}+\mathbf{k} $ , $ \overrightarrow{CA}=2\mathbf{i}-\mathbf{j}-\mathbf{k} $ Clearly $ |AB|,=,|BC|,=,|CA|,=\sqrt{6} $ So these points are vertices of an equilateral triangle.
BETA
