A force of 78 grams acts at the point (2, 3, 5), the direction ratios of the line of action being 2, 2, 1. The magnitude of its moment about the line joining the origin to the point (12, 3, 4) is:
The force has magnitude 78 along direction (2, 2, 1), whose length is 3. So \(\vec{F} = \frac{78}{3}(2, 2, 1) = (52, 52, 26)\).
Moment about the origin: \(\vec{M} = \vec{r} \times \vec{F}\), with \(\vec{r} = (2, 3, 5)\):
\(\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ 52 & 52 & 26 \end{vmatrix} = (78 - 260,\; 260 - 52,\; 104 - 156) = (-182, 208, -52)\)
The unit vector along the line is \(\hat{u} = \frac{(12, 3, 4)}{13}\).
Moment about the line = \(|\vec{M} \cdot \hat{u}| = \frac{|-2184 + 624 - 208|}{13} = \frac{1768}{13} = 136\)
Answer: (B) 136
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