If vec{P} cdot vec{Q} = 0 and vec{P} times ve | vedclass

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(C) Given: $\vec{P} \cdot \vec{Q} = 0$. This implies that $\vec{P}$ and $\vec{Q}$ are perpendicular to each other,so the angle $	heta = 90^{\circ}$.A

Vedclass · Accepted Answer

$90$

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(C) Given: $\vec{P} \cdot \vec{Q} = 0$. This implies that $\vec{P}$ and $\vec{Q}$ are perpendicular to each other,so the angle $	heta = 90^{\circ}$.Also given: $|\vec{P} 	imes \vec{Q}| = PQ$. The magnitude of the cross product is given by $|\vec{P} 	imes \vec{Q}| = PQ \sin 	heta$.Substituting the value of $	heta = 90^{\circ}$,we get $|\vec{P} 	imes \vec{Q}| = PQ \sin(90^{\circ}) = PQ(1) = PQ$.Since both conditions are satisfied at $	heta = 90^{\circ}$,the angle between $\vec{P}$ and $\vec{Q}$ is $90^{\circ}$.

If $\vec{P} \cdot \vec{Q} = 0$ and $\vec{P} \times \vec{Q} = PQ$,then what is the angle between $\vec{P}$ and $\vec{Q}$ (in $^{\circ}$)?

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