Let O be the origin,and overline{OX}, overlin | vedclass

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(D) $(1)$ Since $\overline{OX}, \overline{OY}$ are unit vectors along $QR$ and $RP$,the angle between them is $\pi - R$.Thus,$|\overline{OX} 	imes \o

Vedclass · Accepted Answer

$A, B$

Vedclass · Answer

(D) $(1)$ Since $\overline{OX}, \overline{OY}$ are unit vectors along $QR$ and $RP$,the angle between them is $\pi - R$.Thus,$|\overline{OX} 	imes \overline{OY}| = |\overline{OX}| |\overline{OY}| \sin(\pi - R) = 1 \cdot 1 \cdot \sin R = \sin R$.Since $P+Q+R = \pi$,$\sin R = \sin(\pi - (P+Q)) = \sin(P+Q)$.Thus,option $A$ is correct.$(2)$ We need to find the minimum value of $\cos(P+Q) + \cos(Q+R) + \cos(R+P)$.Since $P+Q+R = \pi$,this is equivalent to $\cos(\pi-R) + \cos(\pi-P) + \cos(\pi-Q) = -(\cos P + \cos Q + \cos R)$.For any triangle,$\cos P + \cos Q + \cos R \leq \frac{3}{2}$.Therefore,$-(\cos P + \cos Q + \cos R) \geq -\frac{3}{2}$.The minimum value is $-\frac{3}{2}$,which occurs for an equilateral triangle.Thus,option $B$ is correct.Combining both,the correct choice is $A, B$.

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