MCQ
If $\vec{P} \cdot \vec{Q} \quad P Q$, then angle between $\vec{P}$ and $\vec{Q}$ is
- ✓$0^{\circ}$
- B$30^{\circ}$
- C$45^{\circ}$
- D$60^{\circ}$
✓
Answer
Correct option: A.
$0^{\circ}$
(a) $\cos \theta=\frac{\vec{P} \cdot \vec{Q}}{ PQ }=1 \therefore \theta=0^{\circ}$
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