What is the angle between vec{A} and vec{B} i | vedclass

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(B) The area of a parallelogram formed by two vectors $\vec{A}$ and $\vec{B}$ as adjacent sides is given by the magnitude of their cross product: $	e

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(B) The area of a parallelogram formed by two vectors $\vec{A}$ and $\vec{B}$ as adjacent sides is given by the magnitude of their cross product: $	ext{Area} = |\vec{A} 	imes \vec{B}|$.Given that $	ext{Area} = \frac{AB}{2}$,we have $|\vec{A} 	imes \vec{B}| = \frac{AB}{2}$.Since $|\vec{A} 	imes \vec{B}| = AB \sin 	heta$,where $	heta$ is the angle between the vectors,we can write: $AB \sin 	heta = \frac{AB}{2}$.Dividing both sides by $AB$,we get $\sin 	heta = \frac{1}{2}$.Therefore,$	heta = 30^{\circ}$ or $	heta = \frac{\pi}{6}$ radians.

Column $I$	Column $II$
$(A) (A+B)$	$(p)$ North-east
$(B) (A-B)$	$(q)$ Vertically upwards
$(C) (A \times B)$	$(r)$ Vertically downwards
$(D) (A \times B) \times (A \times B)$	$(s)$ None

What is the angle between $\vec{A}$ and $\vec{B}$ if $\vec{A}$ and $\vec{B}$ are the adjacent sides of a parallelogram drawn from a common point and the area of the parallelogram is $\frac{AB}{2}$?

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