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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
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The correct answer is:
Obtuse angled
$\overrightarrow{A B}=\text { Position vector of } \vec{B} \text { - Position vector of } \vec{A}=(2 \mathbf{i}+3 \mathbf{j}-6 \mathbf{k})-(6 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})=-4 \mathbf{i}+5 \mathbf{j}-9 \mathbf{k}$
$\Rightarrow|\overrightarrow{A B}|=\sqrt{16+25+81}=\sqrt{122}, \overrightarrow{B C}=\mathbf{i}+3 \mathbf{j}+4 \mathbf{k}$
$\Rightarrow|\overrightarrow{B C}|=\sqrt{1+9+16}=\sqrt{26} \text { and } \overrightarrow{A C}=-3 \mathbf{i}+8 \mathbf{j}-5 \mathbf{k}$
$\Rightarrow|\overrightarrow{A C}|=\sqrt{98}$
$\text {Therefore, } A B^2=122, B C^2=26 \text { and } A C^2=98 .$
$\Rightarrow A B^2+B C^2=26+122=148$
Therefore, $A B^2=122, B C^2=26$ and $A C^2=98$.
Since $A C^2 \lt A B^2+B C^2$, therefore $\triangle A B C$ is an obtuse-angled triangle.
$\Rightarrow|\overrightarrow{A B}|=\sqrt{16+25+81}=\sqrt{122}, \overrightarrow{B C}=\mathbf{i}+3 \mathbf{j}+4 \mathbf{k}$
$\Rightarrow|\overrightarrow{B C}|=\sqrt{1+9+16}=\sqrt{26} \text { and } \overrightarrow{A C}=-3 \mathbf{i}+8 \mathbf{j}-5 \mathbf{k}$
$\Rightarrow|\overrightarrow{A C}|=\sqrt{98}$
$\text {Therefore, } A B^2=122, B C^2=26 \text { and } A C^2=98 .$
$\Rightarrow A B^2+B C^2=26+122=148$
Therefore, $A B^2=122, B C^2=26$ and $A C^2=98$.
Since $A C^2 \lt A B^2+B C^2$, therefore $\triangle A B C$ is an obtuse-angled triangle.
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