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Let the position vectors of two points $A$ and $B$ be $\mathbf{a}+\mathbf{b}+\mathbf{c}$ and $\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}$, respectively. If the points $P$ and $Q$ divide $A B$ in the ratio $1: 3$ internally and externally respectively, then 3|AB $\mid=$
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Verified Answer
The correct answer is:
$4|P Q|$
We have,
$$
\begin{aligned}
\mathbf{O A} & =\mathbf{a}+\mathbf{b}+\mathbf{c}, \mathbf{O B}=\mathbf{a}-2 \mathbf{b}+3 \mathbf{c} \\
\mathbf{A B} & =(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})-(\mathbf{a}+\mathbf{b}+\mathbf{c}) \\
& =-3 \mathbf{b}+2 \mathbf{c}
\end{aligned}
$$
Point $P$ divide $\mathbf{A B}$ in the ratio $1: 3$ internally,
$$
\therefore \mathbf{O P}=\frac{\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}+3 \mathbf{a}+3 \mathbf{b}+3 \mathbf{c}}{\mathrm{l}+3}=\frac{4 \mathbf{a}+\mathbf{b}+6 \mathbf{c}}{4}
$$
Point $Q$ divides $\mathbf{A B}$ in the ratio $1: 3$ externally,
$$
\begin{aligned}
& \therefore \quad \text { OQ }=\frac{\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}-3 \mathbf{a}-3 \mathbf{b}-3 \mathbf{c}}{1-3}=\frac{2 \mathbf{a}+5 \mathbf{b}}{2} \\
& \quad \mathbf{P Q}=\left(\frac{2 \mathbf{a}+5 \mathbf{b}}{2}\right)-\left(\frac{4 \mathbf{a}+\mathbf{b}+6 \mathbf{c}}{4}\right)=\frac{9 \mathbf{b}-6 \mathbf{c}}{4} \\
& |\mathbf{A B}|=\sqrt{9 \mathbf{b}^2+4 \mathbf{c}^2} \text { and }|\mathbf{P Q}|=\frac{1}{4} \sqrt{81 \mathbf{b}^2+36 \mathbf{c}^2} \\
& |\mathbf{P Q}|=\frac{1}{4} \sqrt{9\left(9 \mathbf{b}^2+4 \mathbf{c}^2\right)} \Rightarrow|\mathbf{P Q}|=\frac{3}{4} \sqrt{9 \mathbf{b}^2+4 \mathbf{c}^2} \\
& |\mathbf{P Q}|=\frac{3}{4}|\mathbf{A B}| \Rightarrow 4|\mathbf{P Q}|=3|\mathbf{A B}|
\end{aligned}
$$
$$
\begin{aligned}
\mathbf{O A} & =\mathbf{a}+\mathbf{b}+\mathbf{c}, \mathbf{O B}=\mathbf{a}-2 \mathbf{b}+3 \mathbf{c} \\
\mathbf{A B} & =(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})-(\mathbf{a}+\mathbf{b}+\mathbf{c}) \\
& =-3 \mathbf{b}+2 \mathbf{c}
\end{aligned}
$$
Point $P$ divide $\mathbf{A B}$ in the ratio $1: 3$ internally,
$$
\therefore \mathbf{O P}=\frac{\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}+3 \mathbf{a}+3 \mathbf{b}+3 \mathbf{c}}{\mathrm{l}+3}=\frac{4 \mathbf{a}+\mathbf{b}+6 \mathbf{c}}{4}
$$
Point $Q$ divides $\mathbf{A B}$ in the ratio $1: 3$ externally,
$$
\begin{aligned}
& \therefore \quad \text { OQ }=\frac{\mathbf{a}-2 \mathbf{b}+3 \mathbf{c}-3 \mathbf{a}-3 \mathbf{b}-3 \mathbf{c}}{1-3}=\frac{2 \mathbf{a}+5 \mathbf{b}}{2} \\
& \quad \mathbf{P Q}=\left(\frac{2 \mathbf{a}+5 \mathbf{b}}{2}\right)-\left(\frac{4 \mathbf{a}+\mathbf{b}+6 \mathbf{c}}{4}\right)=\frac{9 \mathbf{b}-6 \mathbf{c}}{4} \\
& |\mathbf{A B}|=\sqrt{9 \mathbf{b}^2+4 \mathbf{c}^2} \text { and }|\mathbf{P Q}|=\frac{1}{4} \sqrt{81 \mathbf{b}^2+36 \mathbf{c}^2} \\
& |\mathbf{P Q}|=\frac{1}{4} \sqrt{9\left(9 \mathbf{b}^2+4 \mathbf{c}^2\right)} \Rightarrow|\mathbf{P Q}|=\frac{3}{4} \sqrt{9 \mathbf{b}^2+4 \mathbf{c}^2} \\
& |\mathbf{P Q}|=\frac{3}{4}|\mathbf{A B}| \Rightarrow 4|\mathbf{P Q}|=3|\mathbf{A B}|
\end{aligned}
$$
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