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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Force vector: \vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}
• Point about which torque is measured: (2,\,3,\,4)
• The force is applied at the intersection of the plane x = 2 and the x -axis, which is the point (2,\,0,\,0) .
Step 2: Determine the Position Vector \vec{r}
The position vector \vec{r} is drawn from the point about which we measure torque (i.e., (2,\,3,\,4) ) to the point of application of the force (i.e., (2,\,0,\,0) ). Mathematically,
\vec{r} = \bigl[(2-2)\hat{i} + (0 - 3)\hat{j} + (0 - 4)\hat{k}\bigr] = 0\,\hat{i} - 3\,\hat{j} - 4\,\hat{k}.
Step 3: Compute the Torque \vec{\tau} Using the Cross Product
Torque is given by the vector cross product:
\vec{\tau} = \vec{r} \times \vec{F}.
Substituting \vec{r} = (0\,\hat{i} - 3\,\hat{j} - 4\,\hat{k}) and \vec{F} = (4\,\hat{i} + 3\,\hat{j} + 4\,\hat{k}) :
\vec{\tau}
= \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
0 & -3 & -4 \\
4 & 3 & 4
\end{vmatrix}.
Expanding this determinant:
\vec{\tau}
= \hat{i}\bigl((-3)\cdot 4 - (-4)\cdot 3\bigr)
- \hat{j}\bigl(0\cdot 4 - (-4)\cdot 4\bigr)
+ \hat{k}\bigl(0\cdot 3 - (-3)\cdot 4\bigr).
Simplify each component:
• \hat{i} component: (-3 \times 4) - (-4 \times 3) = -12 + 12 = 0
• \hat{j} component: -\bigl(0 - (-16)\bigr) = -(16) = -16
• \hat{k} component: (0) - (-12) = 12
Therefore,
\vec{\tau} = 0\,\hat{i} - 16\,\hat{j} + 12\,\hat{k}.
Step 4: Calculate the Magnitude of the Torque
The magnitude of \vec{\tau} is:
|\vec{\tau}|
= \sqrt{(0)^2 + (-16)^2 + (12)^2}
= \sqrt{256 + 144}
= \sqrt{400}
= 20.
Final Answer
|\vec{\tau}| = 20\ \text{(units)}.
