The sum of two forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$ such that $|\overrightarrow{\mathrm{R}}|=|\overrightarrow{\mathrm{P}}| .$ The angle $\theta$ (in degrees) that the resultant of $2 \overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ will make with $\overrightarrow{\mathrm{Q}}$ is
The sum of two forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$ such that $|\overrightarrow{\mathrm{R}}|=|\overrightarrow{\mathrm{P}}| .$ The angle $\theta$ (in degrees) that the resultant of $2 \overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ will make with $\overrightarrow{\mathrm{Q}}$ is
- [JEE MAIN 2020]
- A
$80$
- B
$90$
- C
$85$
- D
$95$
Similar Questions
Magnitudes of two vector $\overrightarrow A $ and $\overrightarrow B $ are $4$ units and $3$ units respectively. If these vectors are $(i)$ in same direction $(\theta = 0^o).$ $(ii)$ in opposite direction $(\theta = 180^o)$, then give the magnitude of resultant vector.
Magnitudes of two vector $\overrightarrow A $ and $\overrightarrow B $ are $4$ units and $3$ units respectively. If these vectors are $(i)$ in same direction $(\theta = 0^o).$ $(ii)$ in opposite direction $(\theta = 180^o)$, then give the magnitude of resultant vector.
A particle is moving along a circular path with a constant speed of $10\,ms^{-1}.$ What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^{o}$ around the centre of the circle .......... $m/s$
A particle is moving along a circular path with a constant speed of $10\,ms^{-1}.$ What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^{o}$ around the centre of the circle .......... $m/s$
- [JEE MAIN 2019]
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Mark the correct statement :-
Mark the correct statement :-
The angle between vector $\vec{Q}$ and the resultant of $(2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}})$ and $(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}})$ is:
The angle between vector $\vec{Q}$ and the resultant of $(2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}})$ and $(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}})$ is:
- [JEE MAIN 2024]