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The position vector of the points $A, B, C$ are $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}),(3 \mathbf{i}-2 \mathbf{j}+\hat{\mathbf{k}})$ and $(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})$
respectively. These points
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respectively. These points
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Verified Answer
The correct answer is:
are collinear
$\overrightarrow{\mathbf{A B}}=(3-2) \hat{\mathbf{i}}+(-2-1) \hat{\mathbf{j}}+(1+1) \hat{\mathbf{k}}$
$\begin{aligned} &=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \Rightarrow|\overrightarrow{\mathbf{A B}}| &=\sqrt{1+9+4}=\sqrt{14} \\ \overrightarrow{\mathbf{B C}} &=(1-3) \hat{\mathbf{i}}+(4+2) \hat{\mathbf{j}}+(-3-1) \hat{\mathbf{k}} \\ &=-2 \hat{\mathbf{i}}+6 \mathbf{j}-4 \hat{\mathbf{k}} \end{aligned}$
$$
\begin{aligned}
\Rightarrow \quad|\overrightarrow{\mathbf{B C}}| &=\sqrt{4+36+16} \\
&=\sqrt{56}=2 \sqrt{14} \\
\overrightarrow{\mathbf{C A}} &=(2-1) \hat{\mathbf{i}}+(1-4) \hat{\mathbf{j}}+(-1+3) \hat{\mathbf{k}} \\
&=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\Rightarrow \quad|\overrightarrow{\mathbf{C A}}| &=\sqrt{1+9+4}=\sqrt{14}
\end{aligned}
$$
So, $|\overrightarrow{\mathbf{A B}}|+|\overrightarrow{\mathbf{A C}}|=|\overrightarrow{\mathbf{B C}}|$ and angle between
$A B$ and $B C$ is $180^{\circ} .$
$\therefore$ Points $A, B, C$ cannot form an isosceles triangle. Hence, $A, B, C$ are collinear.
$\begin{aligned} &=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \Rightarrow|\overrightarrow{\mathbf{A B}}| &=\sqrt{1+9+4}=\sqrt{14} \\ \overrightarrow{\mathbf{B C}} &=(1-3) \hat{\mathbf{i}}+(4+2) \hat{\mathbf{j}}+(-3-1) \hat{\mathbf{k}} \\ &=-2 \hat{\mathbf{i}}+6 \mathbf{j}-4 \hat{\mathbf{k}} \end{aligned}$
$$
\begin{aligned}
\Rightarrow \quad|\overrightarrow{\mathbf{B C}}| &=\sqrt{4+36+16} \\
&=\sqrt{56}=2 \sqrt{14} \\
\overrightarrow{\mathbf{C A}} &=(2-1) \hat{\mathbf{i}}+(1-4) \hat{\mathbf{j}}+(-1+3) \hat{\mathbf{k}} \\
&=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\Rightarrow \quad|\overrightarrow{\mathbf{C A}}| &=\sqrt{1+9+4}=\sqrt{14}
\end{aligned}
$$
So, $|\overrightarrow{\mathbf{A B}}|+|\overrightarrow{\mathbf{A C}}|=|\overrightarrow{\mathbf{B C}}|$ and angle between
$A B$ and $B C$ is $180^{\circ} .$
$\therefore$ Points $A, B, C$ cannot form an isosceles triangle. Hence, $A, B, C$ are collinear.
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