A) Right angled
B) Obtuse angled
C) Equilteral
D) Isosceles
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, \[A{{B}^{2}}=122\], \[B{{C}^{2}}=26\] and \[A{{C}^{2}}=98\]. \[\Rightarrow A{{B}^{2}}+B{{C}^{2}}=26+122=148\] Since \[A{{C}^{2}}<A{{B}^{2}}+B{{C}^{2}}\], therefore \[\Delta ABC\] is an obtuse-angled triangle.
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