Question

The magnitude of the resultant of two vectors \[\vec P\] and \[\vec Q\]. is R. It is given by ______.

पर्याय

• R = \[\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2\mathrm{PQ}\sin\theta}\]

• R = \[\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2\mathrm{PQ}\cos\theta}\]

• R = \[\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+\mathrm{PQ}\sin\theta}\]

• R = \[\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+\mathrm{PQ}\cos\theta}\]

Answer 1

The magnitude of the resultant of two vectors \[\vec P\] and \[\vec Q\]. is R. It is given by R = \[\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2\mathrm{PQ}\cos\theta}\].

Explanation:

By the parallelogram law, the magnitude of the resultant of two vectors P and Q at angle θ is derived from the diagonal of the parallelogram, giving R = \[\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2\mathrm{PQ}\cos\theta}\].