If P = Q = R and vec{P} + vec{Q} = vec{R},let | vedclass

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(B) Given $P = Q = R$. Case $1$: $\vec{P} + \vec{Q} = \vec{R} \implies \vec{Q} = \vec{R} - \vec{P}$. Squaring both sides: $Q^2 = R^2 + P^2 - 2RP \cos

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$	heta_1 = 	heta_2 / 2$

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(B) Given $P = Q = R$. Case $1$: $\vec{P} + \vec{Q} = \vec{R} \implies \vec{Q} = \vec{R} - \vec{P}$. Squaring both sides: $Q^2 = R^2 + P^2 - 2RP \cos 	heta_1$. Since $P = Q = R$,we have $P^2 = P^2 + P^2 - 2P^2 \cos 	heta_1 \implies P^2 = 2P^2(1 - \cos 	heta_1) \implies 1 = 2 - 2 \cos 	heta_1 \implies 2 \cos 	heta_1 = 1 \implies \cos 	heta_1 = 1/2 \implies 	heta_1 = 60^\circ$. Case $2$: $\vec{P} + \vec{Q} + \vec{R} = \vec{0} \implies \vec{P} + \vec{R} = -\vec{Q}$. Squaring both sides: $P^2 + R^2 + 2PR \cos 	heta_2 = Q^2$. Since $P = Q = R$,we have $P^2 + P^2 + 2P^2 \cos 	heta_2 = P^2 \implies 2P^2 + 2P^2 \cos 	heta_2 = P^2 \implies 2 \cos 	heta_2 = -1 \implies \cos 	heta_2 = -1/2 \implies 	heta_2 = 120^\circ$. Comparing the two,$	heta_2 = 2	heta_1$,which implies $	heta_1 = 	heta_2 / 2$.

Column $I$	Column $II$
$(A) \|A+B\|$	$(p) \frac{\sqrt{3}}{2} x^2$
$(B) \|A-B\|$	$(q) x$
$(C) A \cdot B$	$(r) \sqrt{3} x$
$(D) \|A \times B\|$	$(s) \frac{x^2}{2}$

Column-$I$	Column-$II$
$(1)$ Resultant of two mutually perpendicular vectors	$(a)$ Along the bisector of the angle between them
$(2)$ Direction of $\overrightarrow A \times \overrightarrow B$	$(b)$ Coplanar
	$(c)$ Perpendicular to the plane containing $\overrightarrow A$ and $\overrightarrow B$

If $P = Q = R$ and $\vec{P} + \vec{Q} = \vec{R}$,let $\theta_1$ be the angle between $\vec{P}$ and $\vec{R}$. If $\vec{P} + \vec{Q} + \vec{R} = \vec{0}$,let $\theta_2$ be the angle between $\vec{P}$ and $\vec{R}$. What is the relationship between $\theta_1$ and $\theta_2$?

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Column $I$ Column $II$

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