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If $a, b, c$ are distinct real numbers and $P, Q, R$ are three points whose position vectors are respectively $a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}, b \hat{\mathbf{i}}+c \hat{\mathbf{j}}+a \hat{\mathbf{k}}$ and $c \hat{\mathbf{i}}+a \hat{\mathbf{j}}+b \hat{\mathbf{k}}$, then $\angle Q P R=$
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The correct answer is:
$\frac{\pi}{3}$
Position vector of $P=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$
Position vector of $Q=b \hat{\mathbf{i}}+c \hat{\mathbf{j}}+a \hat{\mathbf{k}}$
Position vector of $R=c \hat{\mathbf{i}}+a \hat{\mathbf{j}}+b \hat{\mathbf{k}}$
Now, $\mathbf{P Q}=\mathrm{PV}$ of $Q-\mathrm{PV}$ of $P$
$=(b \hat{\mathbf{i}}+c \hat{\mathbf{j}}+a \hat{\mathbf{k}})-(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}})$
$=(b-a) \hat{\mathbf{i}}+(c-b) \hat{\mathbf{j}}+(a-c) \hat{\mathbf{k}}$
Similarly, $\mathbf{P R}=(c-a) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(b-c) \hat{\mathbf{k}}$
Now, $\mathbf{P Q} \cdot \mathbf{P R}=[(b-a) \hat{\mathbf{i}}+(c-b) \hat{\mathbf{j}}+(a-c) \hat{\mathbf{k}}]$
$\begin{aligned}
& \cdot[(c-a) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(b-c) \hat{\mathbf{k}}] \\
= & (b-a)(c-a)+(c-b)(a-b)+(a-c)(b-c) \\
= & a^2+b^2+c^2-a b-b c-c a \\
|\mathbf{P Q}|= & \sqrt{(b-a)^2+(c-b)^2+(a-c)^2} \\
|\mathbf{P R}|= & \sqrt{(c-a)^2+(a-b)^2+(b-c)^2}
\end{aligned}$
Now, $\cos \theta=\frac{\mathbf{P Q} \cdot \mathbf{P R}}{|\mathbf{P Q}||\mathbf{P R}|}$
$\begin{aligned}
& =\frac{a^2+b^2+c^2-a b-b c-c a}{(a-b)^2+(b-c)^2+(c-a)^2} \\
& =\frac{a^2+b^2+c^2-a b-b c-c a}{2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a} \\
& =\frac{a^2+b^2+c^2-a b-b c-c a}{2\left(a^2+b^2+c^2-a b-b c-c a\right)} \\
\cos \theta & =\frac{1}{2} \\
\therefore \quad \theta & =\frac{\pi}{3}
\end{aligned}$
Hence, $\angle Q P R=\frac{\pi}{3}$.
Position vector of $Q=b \hat{\mathbf{i}}+c \hat{\mathbf{j}}+a \hat{\mathbf{k}}$
Position vector of $R=c \hat{\mathbf{i}}+a \hat{\mathbf{j}}+b \hat{\mathbf{k}}$
Now, $\mathbf{P Q}=\mathrm{PV}$ of $Q-\mathrm{PV}$ of $P$
$=(b \hat{\mathbf{i}}+c \hat{\mathbf{j}}+a \hat{\mathbf{k}})-(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}})$
$=(b-a) \hat{\mathbf{i}}+(c-b) \hat{\mathbf{j}}+(a-c) \hat{\mathbf{k}}$
Similarly, $\mathbf{P R}=(c-a) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(b-c) \hat{\mathbf{k}}$
Now, $\mathbf{P Q} \cdot \mathbf{P R}=[(b-a) \hat{\mathbf{i}}+(c-b) \hat{\mathbf{j}}+(a-c) \hat{\mathbf{k}}]$
$\begin{aligned}
& \cdot[(c-a) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(b-c) \hat{\mathbf{k}}] \\
= & (b-a)(c-a)+(c-b)(a-b)+(a-c)(b-c) \\
= & a^2+b^2+c^2-a b-b c-c a \\
|\mathbf{P Q}|= & \sqrt{(b-a)^2+(c-b)^2+(a-c)^2} \\
|\mathbf{P R}|= & \sqrt{(c-a)^2+(a-b)^2+(b-c)^2}
\end{aligned}$
Now, $\cos \theta=\frac{\mathbf{P Q} \cdot \mathbf{P R}}{|\mathbf{P Q}||\mathbf{P R}|}$
$\begin{aligned}
& =\frac{a^2+b^2+c^2-a b-b c-c a}{(a-b)^2+(b-c)^2+(c-a)^2} \\
& =\frac{a^2+b^2+c^2-a b-b c-c a}{2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a} \\
& =\frac{a^2+b^2+c^2-a b-b c-c a}{2\left(a^2+b^2+c^2-a b-b c-c a\right)} \\
\cos \theta & =\frac{1}{2} \\
\therefore \quad \theta & =\frac{\pi}{3}
\end{aligned}$
Hence, $\angle Q P R=\frac{\pi}{3}$.
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