In a two dimensional motion of a particle- the particle moves from point A- position vector r1 to point B position vector r2- If the magnitude of these vector are respectively r1 - 3 and r2 - 4 and the angle they make with the x-axis are theta-1 - 75 deg- theta-2 - 15 deg- respectively- then magnitude of the displacement vector is-1- 3 nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -2- 13 nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -3- 5 nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -4- 1

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Byju's Answer

Standard IX

Physics

Distance

In a two dime...

Question

In a two dimensional motion of a particle, the particle moves from point A, position vector r₁ to point B position vector r₂. If the magnitude of these vector are respectively r₁ = 3 and r₂ = 4 and the angle they make with the x-axis are θ₁ = 75°, θ₂ = 15° respectively, then magnitude of the displacement vector is-
(1) $\sqrt{3}$ (2) $\sqrt{13}$ (3) $\sqrt{5}$ (4) $\sqrt{1}$

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Solution

Dear Student,
$l e t t h e o r i g i n i s O T h e n D i s p l a c e m e n t v e c t o r w i l l b e \vec{A B} = \vec{O B} - \vec{O A} = \vec{O B} - (- \vec{A O}) = \vec{O B} + \vec{A O} a n d |\vec{O B} + \vec{A O}| = \sqrt{O B^{2} + A O^{2} + 2 O B . A O \cos (180 - 60)} (h e r e a n g l e ∠ A O B = 60) |\vec{O B} + \vec{A O}| = \sqrt{9 + 16 - 2 \times 3 \times 4 \times \frac{1}{2}} = \sqrt{13}$
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