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 \( 180^\circ \).
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]