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JEE MainPhysicsModerateVector

There are two vectors
OA→&OB→\overset{\rightarrow}{OA}\&\overset{\rightarrow}{OB}. Vector
OA→\overset{\rightarrow}{OA} is inxy plane and of magnitude 5 units and
vector
OB→\overset{\rightarrow}{OB} isx-z plane and of magnitude
2\sqrt{2}units as shown. Theangle between the vectors
OA→&OB→\overset{\rightarrow}{OA}\&\overset{\rightarrow}{OB}isθ=cos⁡−1(3n2)\theta = \cos^{- 1}(\frac{3}{n\sqrt{2}}).
Find n.

Show answer and solution
OA→=5cos53∘i^+5sin53∘j^=3i^+4j^\overset{\rightarrow}{OA} = 5cos53^{\circ}\overset{\hat{}}{i} + 5sin53^{\circ}\overset{\hat{}}{j} = 3\overset{\hat{}}{i} + 4\overset{\hat{}}{j}
OB→=2cos45∘i^+2sin45∘k^=1i^+1k^\overset{\rightarrow}{OB} = \sqrt{2}cos45^{\circ}\overset{\hat{}}{i} + \sqrt{2}sin45^{\circ}\overset{\hat{}}{k} = 1\overset{\hat{}}{i} + 1\overset{\hat{}}{k}
cosθ=OA→⋅OB→∣OA→∣∣OB→∣=3+05⋅2=352cos\theta = \frac{\overset{\rightarrow}{OA} \cdot \overset{\rightarrow}{OB}}{|\overset{\rightarrow}{OA}||\overset{\rightarrow}{OB}|} = \frac{3 + 0}{5 \cdot \sqrt{2}} = \frac{3}{5\sqrt{2}}
⇒θ=cos⁡−1(352)\Rightarrow \theta = \cos^{- 1}(\frac{3}{5\sqrt{2}})

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx

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