The angle between vector $\vec{Q}$ and the resultant of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$…

The angle between vector $\vec{Q}$ and the resultant of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$ is :
  1. $\tan ^{-1} \frac{(2 \vec{Q}-2 \vec{P})}{2 \vec{Q}+2 \vec{P}}$
  2. $0^{\circ}$
  3. $\tan ^{-1}(\mathrm{P} / \mathrm{Q})$
  4. $\tan ^{-1}(2 \mathrm{Q} / \mathrm{P})$

Solution

$\begin{aligned} & \overrightarrow{\mathrm{R}}=(2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}})+(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}}) \\ & \overrightarrow{\mathrm{R}}=4 \overrightarrow{\mathrm{Q}} \end{aligned}$ Angle between $\vec{Q}$ and $\vec{R}$ is zero.

Asked in: JEE Main 2024 (05 Apr Shift 1)

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