This problem requires calculating the magnitude of the moment of a force about a specific line in 3D space. We need to use vector algebra concepts, specifically the cross product and dot product.

Understanding the Key Concepts

• Moment of Force: The moment of a force about a point is a measure of its tendency to cause rotation about that point. It is calculated as the cross product of the position vector (from the point to the point of force application) and the force vector ($\vec{M} = \vec{r} \times \vec{F}$).
• Moment about a Line: The moment of a force about a line is the projection of the moment vector (calculated about any point on the line) onto the unit vector representing the direction of the line. Mathematically, it is $| \vec{M} \cdot \hat{v} |$, where $\vec{M}$ is the moment vector and $\hat{v}$ is the unit vector along the line.
• Vectors: We represent points and directions using vectors. The force vector ($\vec{F}$) is determined by its magnitude and direction. The position vector ($\vec{r}$) connects a reference point (on the line) to the point where the force acts. The line itself is defined by a point and a direction vector.

Step-by-Step Calculation

1. Define the Force Vector ($\vec{F}$)

The force has a magnitude of 78 units (assuming grams refers to the magnitude in this context) and its line of action has direction ratios 2, 2, 1.

First, find the unit vector ($\hat{u}$) in the direction of the force:

The direction vector is $<2, 2, 1>$. Its magnitude is $ \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $.

So, the unit vector $\hat{u} = \frac{<2, 2, 1>}{3}$.

The force vector $\vec{F}$ is the magnitude multiplied by the unit direction vector:

$ \vec{F} = 78 \times \hat{u} = 78 \times \frac{1}{3} <2, 2, 1> = 26 <2, 2, 1> = <52, 52, 26> $.

2. Define the Position Vector ($\vec{r}$)

The force acts at point $P = (2, 3, 5)$. The line (axis of rotation) passes through the origin $O = (0, 0, 0)$. The position vector $\vec{r}$ is the vector from the origin to the point P:

$ \vec{r} = \vec{OP} = <2, 3, 5> $.

3. Calculate the Moment Vector ($\vec{M}$)

The moment vector about the origin is calculated using the cross product:

$ \vec{M} = \vec{r} \times \vec{F} $

$ \vec{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 5 \\ 52 & 52 & 26 \end{vmatrix} $

Expanding the determinant:

$ \vec{M} = \mathbf{i}(3 \times 26 - 5 \times 52) - \mathbf{j}(2 \times 26 - 5 \times 52) + \mathbf{k}(2 \times 52 - 3 \times 52) $

$ \vec{M} = \mathbf{i}(78 - 260) - \mathbf{j}(52 - 260) + \mathbf{k}(104 - 156) $

$ \vec{M} = \mathbf{i}(-182) - \mathbf{j}(-208) + \mathbf{k}(-52) $

$ \vec{M} = <-182, 208, -52> $.

4. Define the Unit Vector of the Axis ($\hat{v}$)

The axis of rotation is the line joining the origin $(0, 0, 0)$ to the point $(12, 3, 4)$.

The direction vector of the line is $\vec{v} = <12, 3, 4>$.

The magnitude of $\vec{v}$ is $ |\vec{v}| = \sqrt{12^2 + 3^2 + 4^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = 13 $.

The unit vector $\hat{v}$ along the axis is:

$ \hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{<12, 3, 4>}{13} = <\frac{12}{13}, \frac{3}{13}, \frac{4}{13}> $.

5. Calculate the Magnitude of the Moment about the Line

This is found by taking the scalar product (dot product) of the moment vector $\vec{M}$ and the unit vector of the axis $\hat{v}$.

Magnitude $= | \vec{M} \cdot \hat{v} | $

Magnitude $= | (<-182, 208, -52>) \cdot (<\frac{12}{13}, \frac{3}{13}, \frac{4}{13}>) | $

Magnitude $= | \frac{-182 \times 12}{13} + \frac{208 \times 3}{13} + \frac{-52 \times 4}{13} | $

Simplify the fractions (note that $182 = 13 \times 14$, $208 = 13 \times 16$, $52 = 13 \times 4$):

Magnitude $= | (-14 \times 12) + (16 \times 3) + (-4 \times 4) | $

Magnitude $= | -168 + 48 - 16 | $

Magnitude $= | -120 - 16 | $

Magnitude $= | -136 | $

Magnitude $= 136 $.

Conclusion

The magnitude of the moment of the given force about the specified line is 136 units.