JEE Mains · Physics · STD 11 - 3.1 vectors
The magnitude of vectors \(\overrightarrow{ OA }, \overrightarrow{ OB }\) and \(\overrightarrow{ OC }\) in the given figure are equal. The direction of \(\overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }\) with \(x\)-axis will be
- A \(\tan ^{-1} \frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}\)
- B \(\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1+\sqrt{3}-\sqrt{2})}\)
- C \(\tan ^{-1} \frac{(\sqrt{3}-1+\sqrt{2})}{(1-\sqrt{3}+\sqrt{2})}\)
- D \(\tan ^{-1} \frac{(1+\sqrt{3}-\sqrt{2})}{(1-\sqrt{3}-\sqrt{2})}\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1} \frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}\)
Step-by-step Solution
Detailed explanation
Let magnitude be equal to \(\lambda\). \(\overrightarrow{ OA }=\lambda\left[\cos 30^{\circ} \hat{ i }+\sin 30 \hat{ j }\right]=\lambda\left[\frac{\sqrt{3}}{2} \hat{ i }+\frac{1}{2} \hat{ j }\right]\)…
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