3 forces are applied to a vertex of a cube which are 1, 2 and 3 in magnitude and are directed along the diagonals of the faces of the cube meeting in that vertex. The magnitude of the resultant of these forces is

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answer is C.

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Detailed Solution

Let the diagonals of the faces of the cube are $\overline{i} + \overline{j}, \overline{j} + \overline{k}, \overline{k} + \overline{i}$

$R e s u l \tan t o f t h e t h r e e f o r c e s a l o n g t h e d i a g o n a l s = |\frac{\overline{i} + \overline{j}}{\sqrt{2}} + 2 (\frac{\overline{j} + \overline{k}}{\sqrt{2}}) + 3 (\frac{\overline{k} + \overline{i}}{\sqrt{2}})| = |\frac{4 \hat{i} + 3 \hat{j} + 5 \hat{k}}{\sqrt{2}}| = \frac{\sqrt{16 + 9 + 25}}{\sqrt{2}} = 5$

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