If a, b, c are distinct real numbers and P, Q | vedclass

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(C) The position vectors of points $P, Q, R$ are given by:$\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$$\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$$\

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$\frac{\pi}{3}$

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(C) The position vectors of points $P, Q, R$ are given by:$\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$$\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$$\vec{r} = c \hat{i} + a \hat{j} + b \hat{k}$We need to find $\angle QPR$, which is the angle between vectors $\vec{PQ}$ and $\vec{PR}$.$\vec{PQ} = \vec{q} - \vec{p} = (b-a) \hat{i} + (c-b) \hat{j} + (a-c) \hat{k}$$\vec{PR} = \vec{r} - \vec{p} = (c-a) \hat{i} + (a-b) \hat{j} + (b-c) \hat{k}$The dot product $\vec{PQ} \cdot \vec{PR} = (b-a)(c-a) + (c-b)(a-b) + (a-c)(b-c)$$= (bc - ab - ac + a^2) + (ac - bc - ab + b^2) + (ab - ac - bc + c^2)$$= a^2 + b^2 + c^2 - ab - bc - ca$The magnitudes are:$|\vec{PQ}|^2 = (b-a)^2 + (c-b)^2 + (a-c)^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca)$$|\vec{PR}|^2 = (c-a)^2 + (a-b)^2 + (b-c)^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca)$Thus, $\cos 	heta = \frac{\vec{PQ} \cdot \vec{PR}}{|\vec{PQ}| |\vec{PR}|} = \frac{a^2 + b^2 + c^2 - ab - bc - ca}{2(a^2 + b^2 + c^2 - ab - bc - ca)} = \frac{1}{2}$Therefore, $	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

If $a, b, c$ are distinct real numbers and $P, Q, R$ are three points whose position vectors are respectively $a \hat{i}+b \hat{j}+c \hat{k}$, $b \hat{i}+c \hat{j}+a \hat{k}$ and $c \hat{i}+a \hat{j}+b \hat{k}$, then $\angle Q P R=$

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