Vector Algebra Question 311
Question: If the vectors $ 6\mathbf{i}-2\mathbf{j}+3\mathbf{k},2\mathbf{i}+3\mathbf{j}-6\mathbf{k} $ and $ 3\mathbf{i}+6\mathbf{j}-2\mathbf{k} $ form a triangle, then it is
[Karnataka CET 1999]
Options:
A) Right angled
B) Obtuse angled
C) Equilteral
D) Isosceles
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \overrightarrow{AB} $ = Position vector of $ \overrightarrow{B} $ ? Position vector of $ \overrightarrow{A} $ $ =(2\mathbf{i}+3\mathbf{j}-6\mathbf{k})-(6,\mathbf{i}-2,\mathbf{j}+3\mathbf{k}) $ $ =-4\mathbf{i}+5\mathbf{j}-9\mathbf{k} $
Þ $ |\overrightarrow{AB}|,=\sqrt{16+25+81} $ $ =\sqrt{122} $ , $ \overrightarrow{BC}=\mathbf{i}+3\mathbf{j}+4\mathbf{k} $
Þ $ |\overrightarrow{BC}|,=\sqrt{1+9+16} $ $ =\sqrt{26} $ and $ \overrightarrow{AC}=-3\mathbf{i}+8\mathbf{j}-5\mathbf{k} $
Þ $ |\overrightarrow{AC}|,=\sqrt{98} $ Therefore, $ AB^{2}=122 $ , $ BC^{2}=26 $ and $ AC^{2}=98 $ .
$ \Rightarrow AB^{2}+BC^{2}=26+122=148 $ Since $ AC^{2}<AB^{2}+BC^{2} $ , therefore $ \Delta ABC $ is an obtuse-angled triangle.
BETA
