JEE Mains · Physics · STD 11 - 3.1 vectors
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 _______.
- A \(0^{\circ}\)
- B \(\tan ^{-1} \frac{(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}})}{2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}}}\)
- C \(\tan ^{-1}\left(\frac{P}{Q}\right)\)
- D \(\tan ^{-1}\left(\frac{2 Q}{P}\right)\)
Answer & Solution
Correct Answer
(A) \(0^{\circ}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{R}}=(2 \overrightarrow{\mathrm{Q}}+2 \overrightarrow{\mathrm{P}})+(2 \overrightarrow{\mathrm{Q}}-2 \overrightarrow{\mathrm{P}})\) \(\overrightarrow{\mathrm{R}}=4 \overrightarrow{\mathrm{Q}}\) Angle between \(\vec{Q}\) and \(\vec{R}\) is zero Option…
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