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Standard XII

Physics

Vector Addition

Vectors A and...

Question

Vectors $\to A$ and $\to B$ makes angle of $20^{\circ}$ and $110^{\circ}$ respectively with $+ x$ -axis. The magnitude of these vectors are $5 m$ and $12 m$ respectively. Then the magnitude of the resultant vector is

17 m

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7 m

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13 m

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15 m

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Solution

The correct option is C $13 m$
From diagram it can be seen that, the angle between $\to A$ and $\to B$ is $90^{\circ}$

$R$ = $\sqrt{A^{2} + B^{2} + 2 A B cos θ}$
$R$ = $\sqrt{5^{2} + 12^{2} + 2 \times 5 \times 12 \times cos 90^{\circ}}$
$R$ = $\sqrt{5^{2} + 12^{2} + 0} = 13 m$

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Vectors →A and →B makes angle of 20∘ and 110∘ respectively with +x-axis. The magnitude of these vectors are 5 m and 12 m respectively. Then the magnitude of the resultant vector is

The correct option is C 13 mFrom diagram it can be seen that, the angle between →A and →B is 90∘ R=√A2+B2+2ABcosθ R=√52+122+2×5×12×cos90∘ R=√52+122+0=13 m

Vectors $\to A$ and $\to B$ makes angle of $20^{\circ}$ and $110^{\circ}$ respectively with $+ x$ -axis. The magnitude of these vectors are $5 m$ and $12 m$ respectively. Then the magnitude of the resultant vector is

The correct option is C $13 m$
From diagram it can be seen that, the angle between $\to A$ and $\to B$ is $90^{\circ}$

$R$ = $\sqrt{A^{2} + B^{2} + 2 A B cos θ}$
$R$ = $\sqrt{5^{2} + 12^{2} + 2 \times 5 \times 12 \times cos 90^{\circ}}$
$R$ = $\sqrt{5^{2} + 12^{2} + 0} = 13 m$