If theta is the angle between two vectors vec | vedclass

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(A) $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}| \implies AB \cos 	heta = AB \sin 	heta \implies 	an 	heta = 1 \implies 	heta = 45^{\circ}$

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$(A \rightarrow p, B \rightarrow q, C \rightarrow s, D \rightarrow s)$

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(A) $\vec{A} \cdot \vec{B} = |\vec{A} 	imes \vec{B}| \implies AB \cos 	heta = AB \sin 	heta \implies 	an 	heta = 1 \implies 	heta = 45^{\circ}$ or $135^{\circ}$ (if considering magnitude equality). Thus,$A ightarrow p$.$(B)$ $\vec{A} \cdot \vec{B} = B^2 \implies AB \cos 	heta = B^2 \implies A \cos 	heta = B$. If $A=B$,then $\cos 	heta = 1 \implies 	heta = 0^{\circ}$. Thus,$B ightarrow q, r$.$(C)$ $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \implies A^2 + B^2 + 2AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta \implies 4AB \cos 	heta = 0 \implies 	heta = 90^{\circ}$. Thus,$C ightarrow s$.$(D)$ $|\vec{A} 	imes \vec{B}| = AB \implies AB \sin 	heta = AB \implies \sin 	heta = 1 \implies 	heta = 90^{\circ}$. Thus,$D ightarrow s$.

Column $I$	Column $II$
$(A)$ $\vec{A} \cdot \vec{B} = \|\vec{A} \times \vec{B}\|$	$(p)$ $\theta = 45^{\circ}$ or $135^{\circ}$
$(B)$ $\vec{A} \cdot \vec{B} = B^2$	$(q)$ $\theta = 0^{\circ}$
$(C)$ $\|\vec{A} + \vec{B}\| = \|\vec{A} - \vec{B}\|$	$(r)$ $\vec{A} = \vec{B}$
$(D)$ $\|\vec{A} \times \vec{B}\| = AB$	$(s)$ $\theta = 90^{\circ}$

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

Similar Questions

Given that $\overrightarrow{A} + \overrightarrow{B} = \overrightarrow{C}$ and that $\overrightarrow{C}$ is $\perp$ to $\overrightarrow{A}$. Further,if $|\overrightarrow{A}| = |\overrightarrow{C}|$,then what is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$?

Two vectors $A$ and $B$ have equal magnitude $x$. The angle between them is $60^{\circ}$. Match the following two columns:

Column $I$ Column $II$

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$(B) |A-B|$ $(q) x$

$(C) A \cdot B$ $(r) \sqrt{3} x$

$(D) |A \times B|$ $(s) \frac{x^2}{2}$

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