In a two - dimensional motion of a particle, the particle moves from point A, with position vector $\to [r_{1}]$ , to point B, with position vector ${\to r}_{2}$ . If the magnitudes of these vectors are , respectively, $r_{1} = 3$ and $r_{2} = 4$ and the angles they make with the x - axis are $θ_{1} = 75^{o}$ and $θ_{2} = 15^{o}$ respectively, then find the magnitude of the displacement vector.

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\sqrt{13}

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\sqrt{15}

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Solution

The correct option is B $\sqrt{13}$
Given that,

$r_{1} = 3$

$r_{2} = 4$

$θ_{1} = 75^{0}$

$θ_{2} = 15^{0}$

By cosine rule

$z^{2} = r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos (θ_{1} - θ_{2})$

$z^{2} = {(3)}^{2} + {(4)}^{2} - 2 \times 3 \times 4 cos 60^{0}$

$z^{2} = 9 + 16 - 12$

$z = \sqrt{13}$

Hence, the magnitude of the displacement vector is $\sqrt{13}$

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Similar questions

In a two dimensional motion of particle, the particle moves from point Α, position vector ${\to r}_{1}$ to point B, position vector ${\to r}_{2}$ . If the magnitudes of these vectors are, respectively $r_{1} = 3 a n d r_{2} = 4$ , and the angles they make with the x- axis are $θ_{1} = 75^{0} a n d θ_{2} = 15^{0},$ and respectively, then find the magnitude of the displacement vector.