Given vec{A} + vec{B} + vec{C} = vec{0}. The | vedclass

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(D) Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. Let $|A| = |B| = x$. Then $|C| = \sqrt{2}x$.From $\vec{A} + \vec{B} = -\vec{C}$,squaring both sides

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$90^\circ, 135^\circ, 135^\circ$

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(D) Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. Let $|A| = |B| = x$. Then $|C| = \sqrt{2}x$.From $\vec{A} + \vec{B} = -\vec{C}$,squaring both sides: $A^2 + B^2 + 2AB \cos 	heta_1 = C^2$.Substituting values: $x^2 + x^2 + 2x^2 \cos 	heta_1 = (\sqrt{2}x)^2$.$2x^2 + 2x^2 \cos 	heta_1 = 2x^2 \Rightarrow 2x^2 \cos 	heta_1 = 0 \Rightarrow \cos 	heta_1 = 0 \Rightarrow 	heta_1 = 90^\circ$.Now,consider $\vec{B} + \vec{C} = -\vec{A}$. Squaring both sides: $B^2 + C^2 + 2BC \cos 	heta_2 = A^2$.Substituting values: $x^2 + 2x^2 + 2(x)(\sqrt{2}x) \cos 	heta_2 = x^2$.$2x^2 + 2\sqrt{2}x^2 \cos 	heta_2 = 0 \Rightarrow 2\sqrt{2}x^2 \cos 	heta_2 = -2x^2$.$\cos 	heta_2 = -\frac{1}{\sqrt{2}} \Rightarrow 	heta_2 = 135^\circ$.Since $|A| = |B|$,by symmetry,the angle between $\vec{A}$ and $\vec{C}$ is also $	heta_3 = 135^\circ$.Thus,the angles between the vectors are $90^\circ, 135^\circ, 135^\circ$.

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 $\overrightarrow B$

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

Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. The magnitudes of two vectors are equal,and the magnitude of the third vector is $\sqrt{2}$ times that of the other two. What are the angles between the vectors?

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