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The points whose position vectors are $2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k}, 3 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$ and $4 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$ are the vertices of
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Verified Answer
The correct answer is:
equilateral triangle
$$
\begin{aligned}
& \text { Let } \mathbf{a}=2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k}=\mathrm{OA} \\
& \mathbf{b}=3 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}=\mathrm{OB} \\
& \text { and } \mathbf{c}=4 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k}=\mathrm{OC} \\
& A B=O B-O A=i+j-2 k \\
& B C=O C-O B=\mathbf{i}-2 \mathbf{j}+\mathbf{k} \\
& \text { and } \quad C A=O A-O C=-2 i+j+k \\
& \text { Now, } \quad A B=\sqrt{1+1+4}=\sqrt{6} \\
& \mathrm{BC}=\sqrt{1+4+1}=\sqrt{6} \\
& \text { and } \mathrm{CA}=\sqrt{4+1+1}=\sqrt{6} \\
&
\end{aligned}
$$
Since, the length of all three sides are equal.
So, the triangle is an equilateral triangle.
\begin{aligned}
& \text { Let } \mathbf{a}=2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k}=\mathrm{OA} \\
& \mathbf{b}=3 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}=\mathrm{OB} \\
& \text { and } \mathbf{c}=4 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k}=\mathrm{OC} \\
& A B=O B-O A=i+j-2 k \\
& B C=O C-O B=\mathbf{i}-2 \mathbf{j}+\mathbf{k} \\
& \text { and } \quad C A=O A-O C=-2 i+j+k \\
& \text { Now, } \quad A B=\sqrt{1+1+4}=\sqrt{6} \\
& \mathrm{BC}=\sqrt{1+4+1}=\sqrt{6} \\
& \text { and } \mathrm{CA}=\sqrt{4+1+1}=\sqrt{6} \\
&
\end{aligned}
$$
Since, the length of all three sides are equal.
So, the triangle is an equilateral triangle.
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