Mathematics in Physics

JEE Main level, single correct, medium

23 of 121

Question 23 of 121

The magnitude of vectors OA⃗,OB⃗\vec{OA},\vec{OB} and OC⃗\vec{OC} in the given figure are equal. The direction of OA⃗+OB⃗−OC⃗\vec{OA}+\vec{OB}-\vec{OC} with x−x-axis will be:

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Answer: (B) tan⁡−1⁡(1−3−2)(1+3+2){\tan }^{-1}⁡\dfrac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}

From the above diagram, the equations are,
OA⃗=R[cos⁡30∘i^+sin⁡30∘j^]\vec{OA}=R\left[\cos {30}^{^{\circ}}\hat{i}+\sin {30}^{^{\circ}}\hat{j}\right]
OB⃗=R[cos⁡60∘i^−sin⁡60∘j^]\vec{OB}=R\left[\cos {60}^{^{\circ}}\hat{i}-\sin 60^{\circ}\hat{j}\right]
OC⃗=R[−cos⁡45∘i^+sin⁡45∘j^]\vec{OC}=R\left[-\cos 45^{\circ}\hat{i}+\sin 45^{\circ}\hat{j}\right]
⇒OA⃗+OB⃗−OC⃗=R[12+32+12]i^+R[12−32−12]j^\Rightarrow \vec{OA}+\vec{OB}-\vec{OC}=R\left[\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\right]\hat{i}+R\left[\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\right]\hat{j}, angle with the x−x-axis.
⇒tan⁡α=12−32−1212+32+12=1−3−21+3+2\Rightarrow \tan \alpha =\dfrac{\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}}=\dfrac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}
⇒α=tan⁡−1[1−3−21+3+2]\Rightarrow \alpha ={\tan }^{-1}\left[\dfrac{1-\sqrt{3}-\sqrt{2}}{1+\sqrt{3}+\sqrt{2}}\right].

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