Force $3 N$ , $4 N$ and $12 N$ act at a point in mutually perpendicular directions. The magnitude of the resultant force is:

19 N

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13 N

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11 N

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5 N

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Solution

The correct option is B $13 N$
The forces are mutually perpendicular if they are, $_{1} = 3^i N,_{2} = 4^j N,_{3} = 12^k N$
The resultant force, $\to F =_{1} +_{2} +_{3} = 3^i + 4^j + 12^k$
Magnitude , $F = \sqrt{3^{2} + 4^{2} + 12^{2}} = \sqrt{169} = 13 N$

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Force 3N, 4N and 12N act at a point in mutually perpendicular directions. The magnitude of the resultant force is:

The correct option is B 13 NThe forces are mutually perpendicular if they are, →F1=3^i N,→F2=4^j N,→F3=12^k NThe resultant force,→F=→F1+→F2+→F3=3^i+4^j+12^kMagnitude , F=√32+42+122=√169=13 N

Force $3 N$ , $4 N$ and $12 N$ act at a point in mutually perpendicular directions. The magnitude of the resultant force is:

The correct option is B $13 N$
The forces are mutually perpendicular if they are, $_{1} = 3^i N,_{2} = 4^j N,_{3} = 12^k N$
The resultant force, $\to F =_{1} +_{2} +_{3} = 3^i + 4^j + 12^k$
Magnitude , $F = \sqrt{3^{2} + 4^{2} + 12^{2}} = \sqrt{169} = 13 N$