The vector cross product is anti-commutative by definition. This means that for any two vectors \( \vec{P} \) and \( \vec{Q} \):
\( \vec{P} \times \vec{Q} = -(\vec{Q} \times \vec{P}) \).
The problem states that \( \vec{P} \times \vec{Q} = \vec{Q} \times \vec{P} \).
We can substitute the anti-commutative property into the given equation:
\( \vec{P} \times \vec{Q} = -(\vec{P} \times \vec{Q}) \).
Rearranging the terms, we get:
\( 2(\vec{P} \times \vec{Q}) = \vec{0} \).
This implies that the cross product of \( \vec{P} \) and \( \vec{Q} \) must be the zero vector:
\( \vec{P} \times \vec{Q} = \vec{0} \).
The magnitude of the cross product is given by \( |\vec{P} \times \vec{Q}| = |\vec{P}||\vec{Q}|\sin\theta \), where \( \theta \) is the angle between the vectors.
For the cross product to be zero, assuming \( \vec{P} \) and \( \vec{Q} \) are non-zero vectors, we must have \( \sin\theta = 0 \).
The angles for which \( \sin\theta = 0 \) in the range \( 0^\circ \le \theta<360^\circ \) are \( \theta = 0^\circ \) and \( \theta = 180^\circ \).
The problem specifies the range as \( 0^\circ<\theta<360^\circ \).
Therefore, the only possible value for \( \theta \) is \( 180^\circ \).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,