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If the position vectors of the vertices of a triangle are $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$, then the triangle is
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Verified Answer
The correct answer is:
right angled
Let $\mathbf{A}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{B}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ $\mathbf{C}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
$A B=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}, \mathrm{BC}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
and
$\begin{aligned}
& \mathbf{C A}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& a=|\mathbf{B C}|=\sqrt{4+1+1}=\sqrt{6} \\
& b=|\mathbf{C A}|=\sqrt{1+9+25}=\sqrt{35} \\
& c=|\mathbf{A B}|=\sqrt{1+4+36}=\sqrt{41}
\end{aligned}$
Now, $c^2=a^2+b^2$
$\begin{aligned}
\Rightarrow \quad 41 & =6+35 \\
& =41=41
\end{aligned}$
$\therefore$ It is right angled triangle
$A B=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}, \mathrm{BC}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
and
$\begin{aligned}
& \mathbf{C A}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& a=|\mathbf{B C}|=\sqrt{4+1+1}=\sqrt{6} \\
& b=|\mathbf{C A}|=\sqrt{1+9+25}=\sqrt{35} \\
& c=|\mathbf{A B}|=\sqrt{1+4+36}=\sqrt{41}
\end{aligned}$
Now, $c^2=a^2+b^2$
$\begin{aligned}
\Rightarrow \quad 41 & =6+35 \\
& =41=41
\end{aligned}$
$\therefore$ It is right angled triangle
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